An
Exposition
of
Symbolic Logic
with
Kalish-Montague
derivations
Copyright © 2006-13 by Terence Parsons
all rights reserved
Aug 2013
Preface
The system of logic used here is essentially that of Kalish & Montague 1964 and Kalish,
Montague and Mar, Harcourt Brace Jovanovich, 1992. The principle difference is that written
justifications are required for boxing and canceling: 'dd' for a direct derivation, 'id' for an indirect
derivation, etc. This text is written to be used along with the UCLA Logic 2010 software program,
but that program is not mentioned, and the text can be used independently (although you would
want to supplement the exercises).
The system of notation is almost the same as KK&M; major differences are that the signs '∀' and
'∃' are used for the quantifiers, name and operation symbols are the small letters between ‘a’ and
‘h’, and variables are the small letters between ‘i’ and ‘z’.
The exercises are new.
Chapters 1-3 cover pretty much the same material as KM&M except that the rule allowing for the
use of previously proved theorems is now in chapter 2, immediately following the section on
theorems. (Previous versions of this text used the terminology ‘tautological implication’ in section
2.11. This has been changed to ‘tautological validity’ to agree with the logic program.)
Chapters 4-6 include invalidity problems with infinite universes, where one specifies the
interpretation of notation "by description"; e.g. "R(): ≤". These are discussed in the final
section of each chapter, so they may easily be avoided. (They are not currently implemented in
the logic program.)
Chapter 4 covers material from KK&M chapter IV, but without operation symbols. Chapter 4 also
includes material from KK&M chapter VII, namely interchange of equivalents, biconditional
derivations, monadic sentences without quantifier overlay, and prenex form.
Chapter 5 covers identity and operation symbols.
Chapter 6 covers Fregean definite descriptions, as in KK&M chapter VI.
Version Aug 2013 of An Exposition of Symbolic Logic
is a lightly revised version of the August 2012 version of
An Introduction to Symbolic Logic (also known as Terry-Text).
Copyrighted material
Introduction -- 2
Version of Aug 2013
CONTENTS
Chapter One
Sentential Logic with 'if' and 'not'
1 SYMBOLIC NOTATION
2 MEANINGS OF THE SYMBOLIC NOTATION
3 SYMBOLIZATION: TRANSLATING COMPLEX SENTENCES INTO SYMBOLIC NOTATION
4 RULES
5 DIRECT DERIVATIONS
6 CONDITIONAL DERIVATIONS
7 INDIRECT DERIVATIONS
8 SUBDERIVATIONS
9 SHORTCUTS
10 STRATEGY HINTS FOR DERIVATIONS
11 THEOREMS
12 USING PREVIOUSLY PROVED THEOREMS IN DERIVATIONS
Chapter Two
Sentential Logic with 'and', 'or', if-and-only-if'
1 SYMBOLIC NOTATION
2 ENGLISH EQUIVALENTS OF THE CONNECTIVES
3 COMPLEX SENTENCES
4 RULES
5 SOME DERIVATIONS USING RULES S, ADJ, CB
6 ABBREVIATING DERIVATIONS
7 USING THEOREMS AS RULES
8 DERIVED RULES
9 OFFICIAL CONDITIONS FOR DERIVATIONS
10 TRUTH TABLES AND TAUTOLOGIES
11 TAUTOLOGICAL VALIDITY
Chapter Three
Individual constants, Predicates, Variables and Quantifiers
1 INDIVIDUAL CONSTANTS AND PREDICATES
2 QUANTIFIERS, VARIABLES, AND FORMULAS
3 SCOPE AND BINDING
4 MEANINGS OF THE QUANTIFIERS
5 SYMBOLIZING SENTENCES WITH QUANTIFIERS
6 DERIVATIONS WITH QUANTIFIERS
7 UNIVERSAL DERIVATIONS
8 SOME DERIVATIONS
9 DERIVED RULES
10 INVALIDITIES
11 EXPANSIONS
Copyrighted material
Introduction -- 3
Version of Aug 2013
Chapter Four
Many-Place Predicates
1 MANY-PLACE PREDICATES
2 SYMBOLIZING SENTENCES USING MANY-PLACE PREDICATES
3 DERIVATIONS
4 THE RULE "INTERCHANGE OF EQUIVALENTS"
5 BICONDITIONAL DERIVATIONS
6 SENTENCES WITHOUT OVERLAY OF QUANTIFIERS
7 PRENEX NORMAL FORMS
8 SOME THEOREMS
9 SHOWING INVALIDITY
10 COUNTER-EXAMPLES WITH INFINITE UNIVERSES
Chapter Five
Identity and Operation Symbols
1 IDENTITY
2 AT LEAST AND AT MOST, EXACTLY, AND ONLY
3 DERIVATIONAL RULES FOR IDENTITY
4 INVALIDITIES WITH IDENTITY
5 OPERATION SYMBOLS
6 DERIVATIONS WITH COMPLEX TERMS
7 INVALID ARGUMENTS WITH OPERATION SYMBOLS
8 COUNTER-EXAMPLES WITH INFINITE UNIVERSES
Chapter Six
Definite Descriptions
1 DEFINITE DESCRIPTIONS
2 SYMBOLIZING SENTENCES WITH DEFINITE DESCRIPTIONS
3 DERIVATIONAL RULES FOR DEFINITE DESCRIPTIONS: PROPER DESCRIPTIONS
4 SYMBOLIZING ORDINARY LANGUAGE
5 DERIVATIONAL RULES FOR DEFINITE DESCRIPTIONS: IMPROPER DESCRIPTIONS
6 INVALIDITIES WITH DEFINITE DESCRIPTIONS
7 UNIVERSAL DERIVATIONS
8 COUNTER-EXAMPLES WITH INFINITE UNIVERSES
Copyrighted material
Introduction -- 4
Version of Aug 2013
Introduction
Introduction
Logic is concerned with arguments, good and bad. With the
docile and the reasonable, arguments are sometimes useful in
settling disputes. With the reasonable, this utility attaches only to
good arguments. It is the logician's business to serve the
reasonable. Therefore, in the realm of arguments, it is the logician
who distinguishes good from bad.
Kalish & Montague 1964 p. 1
1 DEDUCTIVE REASONING
Logic is the study of correct reasoning. It is not a study of how this reasoning originates, or what
its effects are in persuading people; it is rather a study of what it is that makes some reasoning
"correct" as opposed to "incorrect". If you have ever found yourself or someone else making a
mistake in reasoning, then this is an example of someone being taken in by incorrect reasoning,
and you have some idea of what we mean by correct reasoning: it is reasoning that contains no
mistakes, persuasive or otherwise.
It is typical in logic to divide reasoning into two kinds: deductive and inductive, or, roughly,
"airtight" and "merely probable". Here is an example of probable reasoning. You have just been
told that Mary bought a new car, and you say to yourself:
In the past, Mary always bought big cars.
Big cars are usually gas-guzzlers.
So she (probably) now has a gas-guzzler.
Your conclusion, that Mary has a gas-guzzler, is not one that you think of as following logically
from the information that you have; it is merely a probable inference.
Inductive Logic, which is the study of probable reasoning, is not very well understood at present.
There are certain rather special cases that are well developed, such as the application of the
probability calculus to gambling games. But a general study has not met with great success.
This is not a book about probable reasoning, but if you are interested in it, this is the place to
start. This is because most studies of Inductive Logic take for granted that you are already
familiar with Deductive Logic -- the logic of "airtight" reasoning -- which forms the subject matter
of this book. So you have to start here anyway.
Here is an example of deductive reasoning. Suppose that you recall reading that either James
Polk or Eli Whitney was a president of the United States, but you can't remember which one.
Some knowledgeable person tells you that Eli Whitney was never president (he was a famous
inventor). Based on this information you conclude that Polk was a president.
The information that you have, and the conclusion that you draw from this information, is:
Either Polk or Whitney was a president.
Whitney was not a president.
So Polk was a president.
Let us compare this reasoning with the other reasoning given above. They both have one thing in
common: the information that you start with is not known for certain. In the first example, you
have only been told that Mary bought a new car, and this may be a lie or a mistake. Likewise,
you may be misremembering her past preferences for car sizes. The same is true in the second
reasoning: you were only told that Eli Whitney was not president -- by someone else or by a
history book -- and your memory that either Polk or Whitney was a president may also be
inaccurate. In both cases the information that you start with is not known for certain, and so in
this sense your conclusions are only probable. Reasoning is always reasoning from some
claims, called the premises of the reasoning, to some further claim, called the conclusion. If the
premises are not known for certain, then no matter how good the reasoning is, the conclusion will
not be known for certain either. (There are certain special exceptions to this; see the exercises
below.) There is, however, a difference in the nature of the inferences in the two cases. In the
Copyrighted material
Introduction -- 5
Version of Aug 2013
Introduction
second piece of reasoning the reasoning itself is airtight in the following sense: If the premises
that you start with are true, then you are guaranteed that your conclusion is true too. That is, if
you were right in thinking that either Polk or Whitney was a president, and if you were right in
thinking that Whitney was not a president, then you must be right in thinking that Polk was a
president. In this case, it is logically impossible for your premises to be true and your conclusion,
nonetheless, to be false. This is an example of what is called validity. If your reasoning is valid,
then, although you are not guaranteed that your conclusion is true, you are guaranteed that it is
true if your premises are.
This guarantee is absent in the case of inductive reasoning. Suppose that Mary has indeed just
bought a new car, and suppose that you are correct in believing that she always bought big cars
in the past, and also correct in believing that big cars are usually gas-guzzlers. You could still be
wrong in your conclusion that she now has a gas-guzzler. Maybe she decided this time to buy a
smaller car. Or maybe she got a big one with some extraordinary new fuel economy equipment.
These may be unlikely, but they are not ruled out, even assuming that all of your premises are
true. The reasoning is not deductively valid because there is a logical possibility that the
conclusion is false even if the premises are all true. In short, in the case of inductive reasoning,
the inconclusiveness of the reasoning itself introduces further uncertainty in addition to the
original uncertainty of the premises.
We rarely have certain knowledge, and a study of logic will not give it to us. Logic is not a
method of achieving certainty in general, though it sometimes yields such knowledge as a
by-product; instead, it is a study of the logical relationships among all our sentences, including
those that are only probable.
2 TRUTH & VALIDITY
A principle unit of investigation in logic is called an argument. An "argument", in its technical
sense, consists of two parts: a set of sentences, called the premises, and a sentence called the
conclusion. The term "argument" may suggest a dispute, but in logic something is called an
argument whether or not any people ever have or ever will disagree about it. Likewise, the
"premises" of such an argument may or may not have been believed or asserted by somebody,
and it is sometimes useful to examine arguments whose "premises" would never be believed by
any rational person. Likewise, by calling something a "conclusion" we do not suggest that
anyone ever has or even should "conclude" this thing on the basis of the premises given. The
point of the terminology is this: a major topic in the study of deductive logic is validity. This is a
relationship between a set of sentences and another sentence; this relationship holds whenever it
is logically impossible for there to be a situation in which all the sentences in the first set are true
and the other sentence false. It turns out to be very useful to study this relationship in complete
generality. That is, it is useful to have a theory which tells us when this relationship holds
between any set of sentences and any other sentence. Since a major practical application of
such a theory is to pieces of reasoning that people actually use, the tradition has arisen of calling
the first set of sentences the "premises", and the other sentence the "conclusion". And since a
practical application of logic is to situations in which people disagree, it is perhaps appropriate to
call the whole thing an "argument". But these are now technical terms. An argument is simply
something that has two parts: a set of sentences called the premises, and another sentence
called the conclusion. For logical purposes, any such combination counts as an argument.
In displaying arguments it is customary to write their premises first, and to indicate the conclusion
by the word like 'so' or a symbol such as '∴'
Either Polk was a president or Whitney was a president.
Whitney was not a president.
∴ Polk was a president.
The triangle made of three dots is an abbreviation of the word `therefore', and is a way of
identifying the conclusion of an argument. In order to save on writing, and also to begin
displaying the form of the arguments under discussion, we will start abbreviating simple
sentences by capital letters. For the time being we will abbreviate Polk was a president by `P',
Copyrighted material
Introduction -- 6
Version of Aug 2013
Introduction
and Whitney was a president by `W'. We will abbreviate Whitney was not a president by `not W'.
So the argument can be shortened to:
P or W
not W
∴P
A major point of this book is to explore the notion of deductive validity. Since the deductive kind
is the only one considered here, we simply refer to it as "validity". In this section we will go over
certain consequences of the following definition of validity:
•
An argument is valid if, and only if, there is no logically possible situation in
which all of its premises are true and its conclusion false.
When we talk about "truth" here we do not have anything deep or mysterious in mind. For
example, we say that the sentence 'There is beer in the refrigerator' is true if there is beer in the
refrigerator, and false if there isn't beer in the refrigerator. That's all there is to it.
We have already seen one case of a valid argument which has all of its premises true and its
conclusion true as well:
P or W
not W
∴P
True
True
True
What other possibilities are there? Well, as we noted above, it is possible to have some of the
premises false and the conclusion false too. (This is sometimes referred to as a case of the
"garbage in, garbage out" principle.) Suppose we use `R' to abbreviate Robert E. Lee was a
president. Then this argument does not have all of its premises true, nor is its conclusion true:
R or W
not W
∴R
False
True
False
Yet this argument is just as good, as far as its validity is concerned, as the first one. If its
premises were true, then that would guarantee that its conclusion would be true too. There is no
logically possible situation in which the premises are all true and the conclusion false. This
argument, though it starts with a false premise and ends up with a false conclusion, has exactly
the same logical form as the first one. This sameness of logical form lies at the foundation of
the theory in this book; it is discussed in the following section.
Although false inputs can lead to false outputs, there is no guarantee that this will happen, for you
can reason validly from false information and accidentally end up with a conclusion that is true.
Here is an example of that:
P or not W
W
∴P
True
False
True
In this example, one of the premises is false, but the conclusion happens to be true anyway.
Mistaken assumptions can sometimes lead to a true conclusion by chance.
The one combination that we cannot have is a valid argument which has all true premises and a
false conclusion. This is in keeping with the definition given above: a deductively valid argument
is one for which it is logically impossible for its conclusion to be false if its premises are all true.
We have seen that there are valid arguments of each of these sorts:
PREMISES
CONCLUSION
all true
true
not all true
false
not all true
true
What about invalid arguments? (That is, what about arguments that are not deductively valid?)
What combination of truth-values can the parts of invalid arguments have? The answer is that
they can have any combination of truth-values whatsoever. Here are some examples:
Copyrighted material
Introduction -- 7
Version of Aug 2013
Introduction
P or W
P
∴W
True
True
False
PREMISES ALL TRUE
CONCLUSION FALSE
P
not W
∴ not R
True
True
True
PREMISES ALL TRUE
CONCLUSION TRUE
P or W
W
∴P
True
False
True
PREMISES NOT ALL TRUE
CONCLUSION TRUE
W or R
P
∴R
False
True
False
PREMISES NOT ALL TRUE
CONCLUSION FALSE
The moral of the story so far is that if you know that an argument is invalid, that fact alone tells
you nothing at all about the actual truth-values possessed by its parts. And if you know that it is
valid, all that that fact tells you about the actual truth-values of its parts is that it does not have all
of its premises true plus its conclusion false.
However, there is more to be said. Suppose that you want to show that an argument is invalid,
but the argument does not already have all true premises and a false conclusion. How can you
do this? One approach is to appeal directly to the characterization of validity, and describe a
possible situation in which the premises are all true and the conclusion false. For example,
suppose someone has given this (invalid) argument:
Either Roosevelt or Truman (or perhaps both) was a president.
Truman was a president.
∴ Roosevelt was a president.
There is no mistake of fact involved here, but the argument is a bad one, and you would like to
establish this. You could do so as follows. You say:
"Suppose that Truman had been a president, but not Roosevelt. In that situation
the premises would have been true, but the conclusion false."
This is enough to show the reasoning bad, that is, to show the argument invalid.
We can do even more than this, as we will see in the next section.
EXERCISES
This book provides a stock of exercises as an aid to learning. They were written in the belief that
the "hands on" approach to modern logical theory is the best way to master it. You will also be
supplied with answers to many of the exercises. You should attempt every exercise on your own,
and then check your efforts against the answers that are given. If you do not understand one
or more of the exercises, ask for help!
Several of the exercises contain material that supplements the explanations in the body of the
text. None of the exercises presuppose material that is not provided in the text or in the exercise
itself.
1. Decide whether each of the following arguments is valid or invalid. If the argument is invalid
then describe a possible situation in which its premises are all true and its conclusion false.
a. Either Polk or Lee was a president.
Either Lee or Whitney was a president.
∴ Either Polk or Whitney was a president.
b. Lee wasn't a president, and Polk was.
Either Polk or Whitney was a president.
∴ Whitney was a president.
Copyrighted material
Introduction -- 8
Version of Aug 2013
Introduction
c. Polk was a president and so was Lee.
Whitney was a president.
∴ Polk was a president and so was Whitney.
d. Either Polk or Whitney was a president.
Lee was not a president.
∴ Lee wasn't a president and Polk was.
2. Which of these are true, and which are false:
a.
b.
c.
d.
e.
f.
g.
Some valid arguments have false conclusions.
No invalid argument has all true premises and a false conclusion.
If an argument is valid, and you produce a new argument from it by adding one or more
premises to it, the resulting argument will still be valid.
If an argument is invalid, and you produce a new argument from it by adding one or more
premises to it, the resulting argument must still be invalid.
If an argument has an impossible premise, it is valid. (An example of an impossible
sentence is `Some giraffes aren't giraffes'.)
If an argument has a necessarily true conclusion, it is valid. (An example of a necessarily
true sentence is `Every giraffe is a giraffe'.)
If an argument has a false premise, it is valid.
3. An argument which is valid and which also has all of its premises true is called sound. Based
on this definition, which of the following are true, and which false:
a.
b.
c.
d.
e.
f.
All valid arguments are sound.
All invalid arguments are unsound.
All sound arguments have true conclusions.
If an argument is sound, and you produce a new argument from it by adding one or more
premises to it, the resulting argument will still be sound.
All unsound arguments are invalid.
If an argument has a necessarily true conclusion, it is sound.
4. Suppose that we have two arguments which are both valid:
W
∴Q
Q
∴S
what do we know about this argument?
W
∴S
5. Suppose that the following two simple arguments are both sound:
W
∴Q
Q
∴S
what do we know about this argument?
W
∴S
6. Suppose that the following argument is invalid:
W
∴S
what do we know about these arguments?
W
∴R
R
∴S
Copyrighted material
Introduction -- 9
Version of Aug 2013
Introduction
7. (a) Give an example of a "reversing" argument, that is, one which is guaranteed to have a
false conclusion if its premises are true, and is guaranteed to have a true conclusion if any of its
premises are false. (b) Give an example of an argument that must have a true conclusion no
matter what the truth-values of its premises. Is this argument valid?
3 LOGICAL FORM
If you want to show that an argument is invalid, you can describe a possible situation in which the
premises are all true and the conclusion false. We illustrated this above with the argument:
Either Roosevelt or Truman (or perhaps both) was a president.
Truman was a president.
∴ Roosevelt was a president.
But this direct appeal to possible situations is sometimes difficult to articulate, and judgments of
possibility can differ. Fortunately, there is another technique that is often more useful. You could
challenge the above reasoning by saying:
"That reasoning is no good. If that reasoning were good,
we could prove that McGovern was a president! For we know that:
Either McGovern or Truman was a president,
and we know:
Truman was a president.
So, by your reasoning we should be able to conclude that
McGovern was a president too!"
This challenge, like the first one, also shows that the argument given above is invalid. But
whereas the first type of challenge focuses on how the ORIGINAL argument works in some
POSSIBLE situation, this second challenge is based on how some OTHER argument works in
the ACTUAL situation. What we do in this second technique is to give an argument that is
different than the first, but closely related to it. In the case in question, the new argument is:
Either McGovern or Truman was a president.
Truman was a president.
∴ McGovern was a president.
We know the new argument is invalid because it actually has all true premises and a false
conclusion (we chose it on purpose to be this way). Since the new argument is invalid, so is the
original one.
But why should the original argument be invalid just because this second argument is invalid?
The answer is that, intuitively speaking, they both employ the same reasoning, and it is the
reasoning that is being assessed when we make a judgment about validity. But how can we tell
that they employ the same reasoning? The answer is that they both have the same form. Each
argument is one in which one of the premises is an "or" statement, with the other premise being
one of the parts of the "or" statement and the conclusion being the other part. This sameness of
structure or form indicates a sameness of the reasoning involved.
A key assumption on which all of modern logical theory is based is that goodness of deductive
reasoning is a matter of form. Any argument which has just the same form as the argument we
were just discussing is invalid, no matter whether its subject matter is religion, politics,
mathematics, or baseball. Likewise, any argument which has this form:
P or W
not W
∴P
Copyrighted material
Introduction -- 10
Version of Aug 2013
Introduction
is valid, regardless of its subject matter. With this in mind, we can give a modern account of
validity due to form:
•
An argument is formally valid if and only if every argument with exactly the same form is
valid.
It follows from this definition that if an argument is formally valid, so is any argument with exactly
that form, if an argument is not formally valid, neither is any argument with exactly that form. A
central preoccupation of modern logic, then, is the investigation and classification of logical forms.
(That is why this logic is called "formal logic".) This will be our business throughout the chapters
that follow.
EXERCISES
1. Decide whether each of the following arguments is valid or invalid. If the argument is invalid
then give an argument which has the same form, and which actually has all true premises and a
false conclusion.
a. Either Polk or Lee was a president.
Either Whitney or Lee was a president.
∴ Either Polk or Whitney was a president.
b. Lee wasn't a president, and Polk was.
Either Polk or Whitney was a president.
∴ Whitney was a president.
c. Polk was a president and so was Lee.
Whitney was a president.
∴ Polk was a president and so was Whitney.
d. Either Polk or Whitney was a president.
Lee was not a president.
∴ Lee wasn't a president and Polk was.
2. Which of these are true, and which are false:
a.
b.
Some invalid arguments have the same forms as valid ones.
You can show an argument valid by producing another argument which has the same
form and which has true premises and a true conclusion.
c.
If you are wondering whether an argument is valid or not, and you fail to find another
argument which has the same form and all true premises and a false conclusion, that
shows the original argument to be valid.
3. Here are some argument forms. For each, say whether every argument with that form is valid.
If it is not valid, give an example of an argument with the given form that has true premises and a
false conclusion.
a.
If A then B
A
∴B
b.
If A then B
B
∴A
c.
not (A and B)
not-B
∴ not A
d.
A or B
B
∴A
Copyrighted material
Introduction -- 11
Version of Aug 2013
Introduction
e.
A and not-A
∴B
f.
A
∴ B or not-B
g.
A or B
not-B or C
∴ A or C
4. Recall that an argument which is valid and which also has all of its premises true is called
sound.
a.
If you are wondering whether an argument is sound, and you manage to find another one
with the same form and having all true premises and a false conclusion, does that show
the original argument to be unsound? Why?
b.
If you are wondering whether an argument is sound, and you manage to find another one
with the same form and having all true premises and a true conclusion, does that show
the original argument to be sound? Why?
c.
If you are wondering whether an argument is sound, and you manage to find another one
with the same form and having all false premises and a false conclusion, does that show
the original argument to be sound? To be unsound? Why?
5. For each of the examples in 3, say whether or not every argument with that form is sound, and
also say whether some argument with that form is sound.
6. {This question is speculative, and does not necessarily have a straightforward answer} Could
there be an argument that is valid but not formally valid? Could there be an argument that is
formally valid but not valid?
4 SYMBOLIC NOTATION
Our investigation of logical forms will take an indirect route, but one that has proved to be
worthwhile. Instead of attempting a direct classification of the logical forms of sentences of
English, we will develop an artificial language that is considerably simpler than English. It will in
some ways be like English without some of the logically irrelevant aspects of English. And it will
lack some of the characteristics that make the use of English confusing when used in
argumentation. For example, the artificial language will lack some of the structural ambiguity of
English. Consider this English sentence:
Mary teaches little girls and boys.
Does this tell us that Mary teaches little girls and little boys, or that she teaches little girls and
regular-size boys? If this sentence occurred in an argument, the validity of the argument might
turn on how the sentence was read. In the artificial language to be developed, structural
ambiguities of this sort will be absent.
The artificial language will be especially designed to make logical form perspicuous. You are
already familiar with this from arithmetic. Consider the partly symbolic sentence:
For any two numbers x and y, x+y = y+x.
It is clear what this says. The same thing can also be said without any symbols:
Given any two numbers, the result of adding them together in one order is the same as
the result of adding them together in the reverse order.
It is apparent that the use of symbols makes the claim clearly and vividly. Our logical symbolism
will be like this.
Copyrighted material
Introduction -- 12
Version of Aug 2013
Introduction
In fact, it will be possible in the symbolic language to tell the logical forms of its sentences just by
examining the shapes and arrangements of its symbols. And it will be possible to evaluate formal
validity and invalidity of arguments expressed in it by a variety of techniques that appeal directly
to the visible arrangements of the symbols in the sentences used.
This does not mean that we will lose sight of reasoning that is expressed in our native tongue.
One of the tasks of learning the artificial language will be to learn how to take sentences of
English and re-express them in the artificial language we are learning. One of our goals, after all,
is to learn about reasoning that we encounter every day, in its natural habitat.
5 IDEALIZATIONS
The data that we have to deal with are incredibly complex, and this is only an introductory text.
So we will idealize from time to time. This is no different from any other art or science. In physics
you usually begin by studying the behavior of bodies falling in a perfectly uniform gravitational
field, or sliding down frictionless planes. There are no perfectly uniform gravitational fields, and
no frictionless planes, yet studying these things gives us clear and simple models that can be
applied to real phenomena as approximations. And then in the advanced courses you can learn
how friction affects the sliding, and how non-uniform fields affect the movement of things in them.
Here are some of the idealizations that we will make in this book: We will look only at arguments
with indicative sentences, not with imperative or interrogative. We will ignore any problems due
to vagueness. For example, given a perfect understanding of the situation, you may still be
unsure whether to say that Mary loves John, because of the vagueness of distinguishing between
loving and liking. We will also totally ignore the fact that sentences may change truth-value over
time and with differing situations. If I say today:
I'm feeling great!
this may be true, but the very same sentence may be false tomorrow. And it may be true when I
say it, yet false when someone else utters it. This "context dependence" of truth has aroused a
great deal of interest, and there are many theories about how it works. They all presuppose that
their readers have already learned the material in this book. We will pretend in our investigations
that sentences come with unique truth-values that do not change with context. The effects of
context constitute an advanced study.
We will also assume that each sentence is either true or false. Again, the question of whether,
and which, sentences lack truth-value is interesting, but is not to be pursued at the beginning.
Many other idealizations will become apparent as we proceed.
6 THE PLAN OF THE TEXT
In this text we will develop a symbolic notation in a step-by-step process. In chapter 1 we
consider simple sentences that are combined together with the "connectives" 'or' and 'if . . .then . .
.'. Even with this very austere notation we can formulate and study a number of formal validities.
In chapter 2 we expand this notation by introducing further connectives: 'and', 'or', and 'if and only
if'. In chapter 3 we vastly expand our symbolic language with the introduction of variables and
quantifiers. Each expansion of the notation builds on what has gone before, so we are
continually increasing our ability to validate formally valid arguments and invalidate arguments
that are not formally valid. Chapters 4-6 contain additional expansions.
Copyrighted material
Introduction -- 13
Version of Aug 2013
Introduction - Answers to the Exercises
Answers to the Exercises -- Introduction
SECTION 2
1. a.
INVALID
Any possible situation in which Lee was a president but neither of the
others was.
b.
INVALID
Any possible situation in which Polk was a president but neither of the
others was.
c.
VALID
d.
INVALID
2. a.
b.
c.
d.
e.
f.
g.
3. a.
b.
c.
d.
e.
f.
Any possible situation in which Whitney was a president and neither of
the others was.
True. (Such an argument will always have at least one false premise.)
False. Some do; some don't.
True.
False. Sometimes adding a premise converts an invalid argument into a valid one, and
sometimes it does not. It depends on what you add.
True. There can't be a possible situation in which it has all true premises and a false
conclusion because there can't be a possible situation in which it has all true premises.
True. There can't be a possible situation in which it has all true premises and a false
conclusion because there can't be a possible situation in which it has a false conclusion.
False. It might be valid, or it might be invalid.
False. Valid arguments with false premises aren't sound.
True. An invalid argument isn't sound because it isn't even valid.
True. The premises are all true, and it's valid, so its conclusion must be true too.
False. If you add a true premise it will remain sound, but if you add a false premise it will
become unsound.
False. A valid argument is unsound if it has a false premise.
False. If the conclusion is necessarily true the argument will be valid, but it still might
have a false premise, and thus be unsound.
4. It has to be valid. For suppose it were not. Then there would be a possible situation in which
A is true and C is false. Since the first argument is valid, B is true in this situation; but then since
the second argument is valid, C is also true in that situation, contradicting our supposition that
there is a situation in which A is true and C is false.
5. It has to be sound. It has to be valid for the same reason as in the previous example. And
since the first argument is sound, A is true. So its premise is true.
6. We know that at least one of them is invalid, but we don't know which. If they were both valid,
the first argument would have to be valid, as in exercise 4. So they aren't both valid. But there
are cases in which the first is valid and the second invalid, and cases in which the first is invalid
and the second valid, and cases in which they are both invalid.
First valid and second invalid:
A
B
C
Polk was a president
Polk or Lee was a president
Lee was a president
First invalid and second valid:
A
B
C
Polk was a president
Polk and Lee were presidents
Lee was a president
Both invalid:
Copyrighted material
Answers to the Exercises -- 14
Version of Aug 2013
Introduction - Answers to the Exercises
A
B
C
7. a.
b.
Polk was a president
Nixon was a president
Lee was a president
Polk was a president.
∴ Polk wasn't a president.
[Naturally, the argument is invalid.]
Polk was a president.
∴ Either Whitney was a president or he wasn't.
This argument is valid; it cannot have all true premises and a false conclusion because it
cannot have a false conclusion.
SECTION 3
1. a.
Either McGovern or Nixon was president.
Either Nixon or Goldwater was president
∴ Either McGovern or Goldwater was president.
b.
The original argument will do; it already has all true premises and a false conclusion.
c.
VALID.
d.
Either Whitney or Polk was a president.
Lee was not a president.
∴ Lee wasn't a president and Whitney was.
2. a.
b.
c.
False.
False. This does not show that no argument with that form has true premises and a false
conclusion.
False. You might not have looked hard enough.
3. a.
VALID
b.
INVALID
If Polk and Lee were both presidents, Polk was a president.
Polk was a president.
∴ Polk and Lee were both presidents.
c.
INVALID
not (Polk was a president and Lee was a president)
not Lee was a president
∴ not Polk was a president
d.
INVALID
Lee or Polk was a president.
Polk was a president.
∴ Lee was a president.
e.
VALID
f.
VALID
g.
VALID
(This depends interpreting `or' inclusively; this is discussed in chapter 2
below.)
4. a.
Yes. It shows the original argument invalid, and an invalid argument is not sound.
b.
No. The original argument could still be invalid, or have a false premise, or both.
Example:
Original argument:
Lee was a president
∴ Whitney was a president
Copyrighted material
"Found" argument:
Nixon was a president.
∴ Kennedy was a president.
Answers to the Exercises -- 15
Version of Aug 2013
Introduction - Answers to the Exercises
c.
It shows neither.
Examples:
Original unsound argument:
Lee wasn't a president
∴ Whitney wasn't a president
Original sound argument:
Lee wasn't a president
∴ Lee wasn't a president
5. a.
b.
c.
d.
e.
f.
g.
6.
"Found" argument:
Nixon wasn't a president.
∴ Kennedy wasn't a president.
"Found" argument:
Nixon wasn't a president.
∴ Nixon wasn't a president.
Some arguments with this form are sound: the ones with true premises.
But not all; some of them have false premises.
None are sound, since none are valid.
None are sound, since none are valid.
But not all; some of them have false premises.
None are sound, since none are valid.
None are sound, since none has a true premise.
Some arguments with this form are sound: the ones with true premises.
But not all; some of them have false premises.
Some arguments with this form are sound: the ones with true premises.
But not all; some of them have false premises.
Many logicians think that there are arguments that are valid, but not formally valid. An
example is:
Herman is a bachelor
∴ Herman is unmarried
The validity of this argument comes from the meaning of the word 'bachelor', and not
from the form of the sentences in the argument.
As we have defined 'formally valid', any argument that is formally valid is automatically
valid.
Copyrighted material
Answers to the Exercises -- 16
Version of Aug 2013
CHAPTER 1 SECTION 1
Chapter One
Sentential Logic with 'if' and 'not'
1 SYMBOLIC NOTATION
In this chapter we begin the study of sentential logic. We start by formulating the basic part of the
symbolic notation mentioned in the Introduction. For purposes of this chapter and the next, our symbolic
sentences will consist entirely of simple sentences, called atomic sentences, together with molecular
sentences made by combining simpler ones with connectives. The simple sentences are capital letters,
which can be thought of as abbreviating sentences of English, as in the Introduction. In this chapter, the
connectives are the negation sign, '~', and the conditional sign, '→'.
The negation sign, '~', is used much as the word 'not' is used in English, to state the opposite of
what a given sentence says. For example, if 'P' abbreviates the sentence 'Polk was a president',
then '~P' abbreviates the sentence 'Polk was not a president'.
The conditional sign, '→', is used much as 'if . . , then . . .' is used in English. If 'P' abbreviates
the sentence 'Polk was a president' and 'W' abbreviates 'Whitney was a president' then '(P→W)'
abbreviates the sentence 'If Polk was a president, then Whitney was a president'.
We need to be precise about exactly what the symbolic sentences of Chapter 1 are:
•
•
•
Any capital letter between 'P' and 'Z' is a symbolic sentence.
(Numerical subscripts may also be used, as in 'P3', 'Q24'.)
If '□' is a symbolic sentence, so is '~□'
If '□' and '○' are symbolic sentences, so is '(□→○)'.
Nothing is a symbolic sentence of Chapter 1 unless it can be constructed
by means of these provisions.
Some terminology:
• A symbolic sentence containing no connectives at all is an atomic sentence. In this chapter and
the next, only sentence letters are atomic.
• Any symbolic sentence that contains one or more connectives is called a molecular sentence.
• We call '~□' the negation of '□'.
• We call any symbolic sentence of the form '(□ → ○)' a conditional sentence; we call '□' the
antecedent of the conditional, and '○' the consequent of the conditional.
Examples of symbolic sentences with minimal complexity are:
U
~U
(U→V)
The first is an atomic sentence. The second is the negation of that atomic sentence. The last is a
conditional whose antecedent is the atomic sentence 'U' and whose consequent is the atomic sentence
'V'.
Once a molecular sentence is constructed, it can itself be combined with others to make more complex
molecular sentences:
~(U→V)
it is not the case that if U then V
(~V → (U→V))
if it is not the case that V then if U then V
~~(V → (U→V))
it is not the case that it is not the case that if V then if U then V
Copyrighted material
Chapter One -- 1
Version of Aug 2013
CHAPTER 1 SECTION 1
The formation rules determine when parentheses occur in a symbolic sentence. When adding a negation
sign to a sentence you do not add any parentheses. These are not symbolic sentences because they
contain extra (prohibited) parentheses :
~(U), ~(~U), ~((U→V))
Although '~(U→V)' has a parenthesis immediately following the negation sign, that parenthesis got into the
sentence when constructing '(U→V)', and not because of the later addition of the negation sign.
When combining sentences with the conditional sign, parentheses are required. For example, this is not a
sentence:
U→V→W
There is one exception to the need for parentheses. If a sentence appears all by itself, not as part of a
larger sentence, then its outer parentheses may be omitted. So these sentences are taken informally to
be conditional symbolic sentences:
U→V
~U → V
(U→V) → ~U
For purposes of Chapter 1:
A sentence is in official notation if it can be constructed by using the processes given in the
box above.
It is in informal notation if it can be put into official notation by enclosing it in a single pair of
parentheses.
Anything that is not in either official notation or informal notation is not a sentence at all.
Any well-formed sentence can be "parsed" into its constituents. You begin with the sentence itself, and
you indicate below it how it is constructed out of its constituents. First you locate the main connective,
which is the last connective introduced when constructing the sentence. If the sentence is a negation, the
main connective is the negation sign; you draw a vertical line under it and write the part of the sentence to
which the negation sign is applied. If it is a conditional, the main connective is the conditional sign; you
draw branching lines below the main conditional sign and write the antecedent and consequent:
~P
|
P
P→Q
2
P
Q
If the parts are themselves complex, the parsing may be continued:
~~(P → Q)
|
~(P→ Q)
|
P→Q
2
P
Q
Copyrighted material
P → ~(Q→R)
2
P
~(Q→R)
|
Q→R
2
Q
R
(P→Q) → (~R→S)
2
P→Q
~R→S
2
2
P
Q ~R
S
|
R
Chapter One -- 2
Version of Aug 2013
CHAPTER 1 SECTION 1
EXERCISES
1. For each of the following state whether it is a sentence in official notation, or a sentence in informal
notation, or not a sentence at all. If it is a sentence, parse it as indicated above.
a.
b.
c.
d.
e.
f.
g.
h.
i.
Copyrighted material
~~~P
~Q→~R
~(Q~→R)
~(~P)→~R
(P→Q) → (R→~Q)
P → (Q→R) → Q
(P → (Q→R) → Q)
(~S→R) → ((~R→S) → ~(~S→R))
P → (Q→P)
Chapter One -- 3
Version of Aug 2013
CHAPTER 1 SECTION 2
2 MEANINGS OF THE SYMBOLIC NOTATION
The negation sign: The logical import of the negation sign is this: it makes a sentence that is false if the
sentence to which it is prefixed is true, and true if the sentence to which it is prefixed is false. It is
common to summarize this behavior of the negation sign by means of what is called a truth table:
□
T
F
~□
F
T
In this table, 'T' stands for 'true' and 'F' for 'false'. Reading across the rows, the table says that when a
sentence '□' is true, its negation, '~□' is false, and when a sentence '□' is false, its negation '~□' is true.
When a connective can be defined by a truth table in this way, the connective is said to be "truth
functional". This means that the truth value of the whole sentence that is created by combining this
connective with another sentence is completely determined by ("is a function of") the truth value of the
sentence with which it is combined. All connectives in our logical notation will be truth functional.
The negation sign corresponds naturally to either of two locutions in English. One is the locution 'it is not
the case that' placed on the front of a sentence. The other is the word 'not' when used within a simple
sentence. So if 'S' is taken to abbreviate 'The salad was tasty', these have the same logical content:
~S
It is not the case that the salad was tasty
The salad was not tasty
In the logical tradition, the locution 'fail to' is often taken to have the meaning of negation. This is limited to
certain uses of English. For example, if you say Samantha failed to reach the summit you are probably
reporting a situation in which she tried to reach the summit, but did not reach it. However, if you say that
the number of buttons on a shirt fail to match the number of buttonholes, you are probably not saying that
the buttons tried to match the buttonholes but did not match them, you may only be saying that the buttons
and buttonholes do not match. In this usage, 'fail to' may report only negation. The sentence:
The salad failed to be poisonous
then reports the negation of the sentence 'The salad was poisonous', and it may be symbolized '~P'. In
the exercises we will assume that 'fail to' only amounts to negation.
The conditional sign: The conditional sign is meant to capture some part of the logical import of 'if . . ,
then' in English. But it is not completely clear under what circumstances an 'if . . , then' claim in English is
true. It seems clear that any English sentence of the form 'If P then Q' is false when 'P' is true but 'Q' is
false. If you say 'If the Angels win there will be a thunderstorm', then if the Angels do win and if there is no
thunderstorm, what you said is false. In other cases things are not so clear. Consider these conditional
sentences uttered in normal circumstances:
If it rains, the game will be called off.
If the cheerleaders are late, the game will be called off.
Now suppose that it rains, and the cheerleaders are late, and the game is called off. Are the sentences
above true or false? Most people would be inclined to say that the first is true. But the second is less
obvious. After all, the game was not called off because the cheerleaders were late. So there is something
funny about the second sentence. If it is false, it will be impossible to capture the logical import of
conditionals by means of any truth functional connective. For the truth of the first sentence above requires
that some conditionals be true when both their parts are true, and the second would require that some
conditionals be false when both their parts are true.
However, you might hold that the second sentence above is true. Granted, the game was not called off
because the cheerleaders were late, but so what? The second sentence doesn't say anything at all about
why the game was called off. It only says that it will be called off if the cheerleaders are late; and they
were late, and the game was called off, so it is true. If so, perhaps conditionals are truth functional.
There is no universal agreement about how conditionals work in natural language. The position taken in
Copyrighted material
Chapter One -- 4
Version of Aug 2013
CHAPTER 1 SECTION 2
this text is that 'if . . , then. . .' is sometimes used to express what is called the "material conditional". This
is the use of 'if . . , then . . .' where a conditional sentence is false in case the antecedent is true and the
consequent false, and it is true in every other case. This use is truth functional. It is described by means
of this truth table:
□
○
□→○
T
T
T
T
F
F
F
T
T
F
F
T
The conditional is used in this way by mathematicians, and by others. We will assume in doing exercises
and examples that the logical import of 'if . . , then' is intended to coincide with our symbolic '→'. There
may be other uses of 'if . . , then' that convey more than '→', but we will not address them in this text.
The word 'if' in English has many synonyms. In at least some contexts these are all interchangeable:
if
provided that
assuming that
given that
in case
on the condition that
If Maria sings, Xavier will leave
Provided that Maria sings, Xavier will leave
Assuming that Maria sings, Xavier will leave
Given that Maria sings, Xavier will leave
In case Maria sings, Xavier will leave
On the condition that Maria sings, Xavier will leave
Using 'S' for 'Maria sings' and 'X' for 'Xavier will leave', these can all be symbolized as
S→X
'If' clauses in English may also occur at the end of a sentence instead of at the beginning. So these also
may be symbolized as 'S→X':
Xavier will leave if Maria sings
Xavier will leave provided that Maria sings
Xavier will leave assuming that Maria sings
Xavier will leave given that Maria sings
Xavier will leave in case Maria sings
Xavier will leave on the condition that Maria sings
In either use, the word 'If' immediately precedes the antecedent of the conditional.
Many of these "synonyms" of 'if' can be used to say more than what is said with a simple use of the word
'if'. For example, a person who says 'assuming that' may want to convey that s/he is indeed making a
certain assumption, and not just saying 'if'. But in other contexts no assuming is indicated. A physicist
who says 'Assuming that there are planets with orbits outside the orbit of Pluto, we will need to send space
probes to investigate them' may simply be responding to the question 'What if there are planets beyond
Pluto?', and not doing any assuming at all. In doing the exercises we will take for granted that the
locutions identified above are being used in the most minimal sense of 'if', which we take to be that of the
connective '→'.
Only if: The word 'only' can be added to the word 'if', to make 'only if'. The 'only' has the effect of
reversing antecedent and consequent. As a result, whereas 'if', when used alone, immediately precedes
the antecedent of a conditional, 'only if' immediately precedes the consequent. So we have these
equivalences:
If P, Q
Only if P, Q
P→Q
Q→P
P if Q
P only if Q
Q→P
P→Q
Some will find it more natural to represent 'P only if Q' by 'If not Q then not P', or '~Q → ~P'. It will turn out
that this is logically equivalent to 'P → Q'. We will generally use the latter form because it's simpler.
Copyrighted material
Chapter One -- 5
Version of Aug 2013
CHAPTER 1 SECTION 2
When ‘only if' comes first, there are grammatical changes in the last clause, which converts into its
interrogative word order:
The game will be called off only if it rains = Only if it rains will the game be called off
'Only' may also precede any of the synonyms of 'if', so these all may be symbolized as 'X → S':
Xavier will leave only if Maria sings
Xavier will leave only provided that Maria sings
Xavier will leave only assuming that Maria sings
Xavier will leave only given that Maria sings
Xavier will leave only in case Maria sings
Xavier will leave only on the condition that Maria sings
Only if Maria sings will Xavier leave
Only provided that Maria sings will Xavier leave
Only assuming that Maria sings will Xavier leave
Only given that Maria sings will Xavier leave
Only in case Maria sings will Xavier leave
Only on the condition that Maria sings Xavier will leave
EXERCISES
For these exercises assume that 'S' abbreviates 'Susan will be late' and 'R' abbreviates 'It will rain'.
1. For each of the following sentences say which symbolic sentence is equivalent to it.
a. Only if it rains will Susan be late
S→R
R→S
b. Susan will be late provided that it rains
S→R
R→S
c. Susan won't be late
~S
~~S
d. Susan will be late only if it rains
S→R
R→S
e. Given that it rains, Susan will be late
S→R
R→S
2. Symbolize each of the following:
a. Susan will be late only provided that it rains
b. Only on condition that it rains will Susan be late
c. Susan will be late only in case it rains
d. Susan will be late only if it rains
e. It is not the case that Susan will be late
Copyrighted material
Chapter One -- 6
Version of Aug 2013
CHAPTER 1 SECTION 3
3 SYMBOLIZATION: TRANSLATING COMPLEX SENTENCES INTO SYMBOLIC NOTATION
In representing simple sentences of English by sentential letters you need to say which letter abbreviates
which sentence. So far this has been done informally, by choosing sentential letters that are already used
in a prominent place in the English sentence, as in using 'S' for 'Sally will be late'. But if different English
sentences share their prominent letters, we can soon run out of natural sentential letters to choose. So
instead we give a scheme of abbreviation, which pairs off sentence letters with the English sentences
that they abbreviate. An example is:
X
Y
Susan will be late
It will rain
With this scheme we would represent 'Given that it rains, Susan will be late' by:
Y→X
Any way of "symbolizing" English sentences in logical notation, or of "translating" English into logical
notation, is done relative to a scheme of abbreviation. A translation or symbolization that is correct on one
scheme may be incorrect on others.
Complex sentences of English generally translate into complex sentences of the logical notation. Here it
is important to be clear about the grouping of clauses in the English sentence. Consider, the sentence:
If Roberta doesn't call, Susan will be distraught
This is a conditional whose antecedent is a negation. Using 'P' for 'Roberta calls' and 'Q' for 'Susan will be
distraught', this may be symbolized:
~P → Q
It is not the negation of a conditional:
~(P→Q)
To make the negation of a conditional, you need to say something like:
It is not the case that if Roberta calls, Susan will be distraught
which is symbolized as:
~(P → Q)
There are a few fundamental principles that govern symbolizations of English sentences in the logical
notation.
SOURCES OF '~'
The locution 'fail to' always yields a negation sign that applies to the symbolization of the
smallest sentence that 'fail to' is part of.
Likewise for the word 'not'.
The expression 'it is not the case that' applies to a sentence immediately following it.
Notice that in the sentence 'It is not the case that Willa will leave if Sam does' there are two grammatical
sentences that immediately follow 'It is not the case that', namely, 'Willa will leave' and 'Willa will leave if
Sam does'. So there are two ways to symbolize the sentence:
S → ~W
~(S→W)
The sentence is in fact ambiguous.
Copyrighted material
Chapter One -- 7
Version of Aug 2013
CHAPTER 1 SECTION 3
SOURCES OF '→'
If: The word 'if' always gives rise to a conditional, '□→○'.
Wherever 'if' occurs (not as part of 'only if'), the antecedent of the conditional is the
symbolization of a sentence immediately following 'if.
The consequent of the conditional is either the symbolization of a sentence immediately
preceding 'if' (with no comma in between) -- as in '○ if □' -- or it is the symbolization of a
sentence immediately following the sentence that is symbolized as the antecedent -- as in
'if □ then ○'.
Then: If 'then' occurs it must be paired with a preceding 'if'.
The antecedent of the conditional introduced by 'if' is the symbolization of the sentence exactly
between 'if' and 'then'.
Its consequent is the symbolization of a sentence immediately following 'then'.
Only if: The expression 'only if' always gives rise to a conditional, '□→○'.
The consequent of '□→○' is the symbolization of a sentence immediately following 'only if' -as in '□ only if ○' -- or as in 'only if ○, □'.
The antecedent of '□→○' is the symbolization of a sentence immediately preceding 'only if'
(with no comma in between) -- '□ only if ○' -- or of a sentence immediately following the
consequent -- as in 'only if ○, □'. In the latter case, that sentence is grammatically changed
(to its interrogative word order).
Illustration: These principles determine that the sentence: 'Pat won't call only if it is not the case that the
quilt is dirty' is symbolized:
~P → ~Q
The (contracted) 'not' in 'Pat won't call' yields a negation that applies directly to 'P'. The 'it is not the case
that' yields a negation that applies directly to 'Q', since 'the quilt is dirty' is the only sentence immediately
following 'it is not the case that'. The only sentence immediately to the left of the 'only if' is 'Pat won't call',
so that is the antecedent of the conditional, and the only sentence immediately to the right of the 'only if' is
'it is not the case that the quilt is dirty', so that is the consequent.
Illustration: In the sentence 'If Wilma leaves then Xavier stays if Yolanda sings' the first 'if . , then . . .'
exactly encloses 'Wilma leaves', so W is the antecedent of the first conditional. There are two sentences
immediately following 'then'; they are the whole 'Xavier stays if Yolanda sings' and just 'Xavier stays'. So
the sentence must have the form:
W → (Xavier stays if Yolanda sings)
or
(W → X) if Yolanda sings
The second 'if' comes between its consequent and antecedent. It must give rise to a conditional that has
'Y' as its antecedent, since the only sentence following the 'if' is 'Yolanda sings'. The consequent of that
conditional can be the symbolization of either just 'Xavier stays', or 'If Wilma leaves then Xavier stays',
since each of these immediately precedes the 'if'. The first of these fits with the first partial symbolization
above, giving:
W → (Y→X)
and the second fits with:
Y → (W→X).
Both of these symbolizations are possible, which agrees with the intuition that the original English
sentence is ambiguous. (Some people find the first reading more natural than the second, but the second
is a possible reading under some circumstances.)
Copyrighted material
Chapter One -- 8
Version of Aug 2013
CHAPTER 1 SECTION 3
Illustration: In the sentence 'If Wilma leaves then Xavier stays only if Yolanda sings' the first 'if' is exactly
as in the previous case, so the sentence has the form:
W → (Xavier stays only if Yolanda sings)
or
(W → X) only if Yolanda sings
The 'only if' comes between its antecedent and consequent. It must give rise to a conditional that has 'Y'
as its consequent, since the only sentence following the 'only if' is 'Yolanda sings'. The antecedent of that
conditional can be the symbolization of either just 'Xavier stays', or 'If Wilma leaves then Xavier stays',
since each of these immediately precedes the 'only if'. The first of these fits with the first partial
symbolization above, giving:
W → (X→Y)
and the second fits with:
(W→X) → Y
Again, the sentence is ambiguous.
The principles above also apply when 'if' is replaced by one of its synonyms, such as 'given that'.
So 'If Wilma leaves then Xavier stays provided that Yolanda sings' has the same symbolization options as
'If Wilma leaves then Xavier stays if Yolanda sings':
W → (Y→X)
and
Y → (W→X).
Commas: We have seen that the fundamental principles governing words that yield negations and
conditionals can permit a significant amount of ambiguity. A common way to eliminate such ambiguity
from sentences is to use commas to indicate how parts of the symbolization are to be grouped. Commas
are used for a wide variety of purposes, so the presence of a comma may be irrelevant to the
symbolization. But sometimes they are used to indicate that sentences should be grouped together, that
is, combined into a single sentence. When this happens, the comma appears right after the sentence that
results from the grouping. Or, a comma may be used to indicate that sentences to the right should be
grouped together.
COMMAS
A comma indicates that the symbolizations of sentences to its left should be combined into a
single sentence, or that sentences to its right should be combined into a single sentence.
For example, above we saw that the sentence:
If Wilma leaves then Xavier stays if Yolanda sings
has two possible symbolizations:
W → (Y→X)
and
Y → (W→X).
If a comma appears before the first 'then':
If Wilma leaves, then Xavier stays if Yolanda sings
this forces the first symbolization, since it requires that 'X' and 'Y' be grouped together. But if the comma
appears later:
Copyrighted material
Chapter One -- 9
Version of Aug 2013
CHAPTER 1 SECTION 3
If Wilma leaves then Xavier stays, if Yolanda sings
this forces the second symbolization, since now 'W' and 'X' must be grouped together.
Sometimes there are no phrases to be kept together, as in:
If Wilma leaves, then Xavier stays
There is only one sentence to the left of the comma, and only one to the right, so the comma is redundant;
we just have:
W→X
EXERCISES
For these questions please use the following scheme of abbreviation:
V
W
Y
Veronica will leave
William will leave
Yolanda will leave
1. For each of the following say which of the proposed translations are correct.
a. If Veronica doesn’t leave William won’t either
~(V→W)
~V→~W
V → ~~W
b. William will leave if Yolanda does, provided that Veronica doesn’t
(W→Y) → ~V
V → (W→Y)
~V → (Y→W)
c. If Yolanda doesn’t leave, then Veronica will leave only if William doesn’t
~Y → (~W → V)
~Y → (V → ~W)
~W → (~Y→V)
d. If Yolanda doesn’t leave then Veronica will leave, given that William doesn’t
~Y → (~W → V)
~Y → (V → ~W)
~W → (~Y→V)
2. For each of the following produce a correct symbolization
a. William will leave if Veronica does
b. Veronica won't leave if William does
c. If Veronica leaves, then if William doesn't leave, Yolanda will leave
d. If Veronica doesn't leave if William doesn't, then Yolanda won't.
e. William won’t leave provided that Veronica doesn’t leave
f. If William leaves, then if Veronica leaves so will Yolanda
g. William will leave only if if Veronica leaves then so will Yolanda
h. William will leave only if Veronica leaves, only provided that Yolanda will leave
Copyrighted material
Chapter One -- 10
Version of Aug 2013
CHAPTER 1 SECTION 4
4 RULES
A "rule" is a particular valid form of argument which may be used in extended reasoning. Certain rules
involving negations and conditionals have been recognized for centuries, and they have traditional names.
The basic rules used in this chapter are:
RULES
Repetition:
□
∴ □
Modus Ponens:
□→○
□
∴○
Modus Tollens:
□→○
~○
∴ ~□
Double negation:
□
∴ ~~□
or
~~□
∴□
Repetition is the most trivial rule; it indicates that if you have any sentence you may validly infer it from
itself. Although trivial, this rule will have an important use later when we construct a method of showing
that an argument is valid.
Modus ponens indicates that if you have any conditional sentence along with its antecedent you may
infer its consequent. For example, this valid argument is an instance of modus ponens:
If Polk was a president, so was Whitney
Polk was a president
∴ Whitney was a president
P→W
P
∴ W
This rule may be justified by noting that a conditional with a true antecedent and false consequent is false.
Modus Tollens indicates that if you have any conditional sentence along with the negation of its
consequent, you may infer the negation of its antecedent. For example, this valid argument is an instance
of modus tollens:
If Polk was a president, so was Whitney
Whitney wasn't a president
∴ Polk wasn't a president
P→W
~W
∴ ~P
This rule may be justified in a similar way to that used in justifying modus ponens.
Double negation indicates that from any sentence you may infer the result of putting two negation signs
on the front, or vice versa. For example, both of these valid arguments are instances of double negation:
Polk was a president
P
∴ ~~P
It is not the case that Polk wasn't a president
~~P
∴ P
∴ It is not the case that Polk wasn't a president
∴ Polk was a president
It should be obvious upon reflection that any argument whose conclusion follows from its premises by a
single application of one of these rules is formally valid. That is, there cannot be a situation in which it has
true premises and a false conclusion.
These rules apply to anything that fits their pattern, even if it is complex. For example, this is an instance
of modus ponens:
Copyrighted material
Chapter One -- 11
Version of Aug 2013
CHAPTER 1 SECTION 4
If Polk was a president, Whitney wasn't
Polk was a president
∴ Whitney wasn't a president
∴
P→~W
P
~W
And so is this:
If Polk was a president then if Roosevelt was a president, so was Truman
Polk was a president
∴ If Roosevelt was a president then so was Truman
P→(R→T)
P
R→T
The following is not an instance of modus ponens:
If Whitney was a president, so was Truman
Truman was a president
∴ Whitney was a president
W→T
T
∴ W
This argument is in fact invalid. It is an instance of a famous fallacy called "affirming the consequent".
Likewise, the following is not an instance of modus tollens:
If Whitney was a president, so was Truman
Whitney wasn't a president
∴ Truman wasn't a president
W→T
~W
∴ ~T
This too is a famous fallacy, called "denying the antecedent".
EXERCISES
1. For each of the following arguments, say whether it is an instance of modus ponens, or modus tollens,
or double negation, or none of the above.
a.
P→~Q
Q
∴ ~P
b.
~P → Q
∴ ~(P → Q)
c.
~~(P→Q)
∴ P→Q
d.
~P → ~Q
~P
∴ ~Q
e.
~P → ~Q
~~Q
∴ ~~P
f.
P→Q
~R
∴ ~P
g.
P → (R→Q)
P
∴ R→Q
h.
P → (R→Q)
R → ~Q
∴ ~P
i.
~~(P → Q)
~Q
∴ P→Q
2. Given the sentences below, say what, if anything, can be inferred in one step by modus ponens, or
modus tollens, or double negation.
a.
~W → ~X
~W
∴ ?
b.
~W → ~X
~~X
∴ ?
c.
W→X
~W
∴ ?
d.
W → (R→X)
W
∴ ?
e.
W → (R→X)
~R → X
∴ ?
f.
~~(W → X)
~X
∴ ?
g.
W→~X
X
∴ ?
h.
~W → X
∴ ?
i.
~~(W→X)
∴ ?
Copyrighted material
Chapter One -- 12
Version of Aug 2013
CHAPTER 1 SECTION 5
5 DIRECT DERIVATIONS
Complex reasoning often consists of stringing together simple inferences so as to show the validity of a
complex argument. Here is an example. We are given the following premises and conclusion:
If Polk was a president then if Whitney was a president so was Trump.
Polk wasn't a president only if Trump was a president.
Trump wasn't a president.
∴ Whitney wasn't a president.
We may reason as follows:
Since Trump wasn't a president (given) and Polk wasn't a president only if Trump was (given), it is
not the case that Polk wasn't a president. So Polk was a president. But if Polk was a president,
then if Whitney was a president, so was Trump (given); so if Whitney was a president so was
Trump. But Trump wasn't a president (given), so Whitney wasn't a president.
The pattern of reasoning is easier to follow if the sentences are given in symbolic form:
First premise:
Second premise:
Third premise:
Conclusion:
P→(W→T)
~P→T
~T
∴ ~W
The above reasoning can be systematized and explained using the following format, where each line
contains a sentence followed by a “justification” – an explanation of why the sentence is there.
1. To show ~W:
"Since Trump wasn't a president (given) and Polk wasn't a president only if Trump was (given), it
is not the case that Polk wasn't a president."
2.
3.
4.
~T
~P→T
~~P
pr
pr
2 3 mt
this line is a premise
this line is a premise
this line follows from 2 and 3 by modus tollens
"So Polk was a president"
5.
P
4 dn
this line follows from line 4 by double negation
"But if Polk was a president, then if Whitney was a president, so was Trump (given); so if Whitney
was a president so was Trump."
6.
7.
P→(W→T)
W→T
pr
5 6 mp
this line is a premise
this line follows from 5 and 6 by modus ponens
"But Trump wasn't a president (given), so Whitney wasn't a president."
8.
9.
~T
~W
pr
7 8 mt
this line is a premise
this line follows from 7 and 8 by modus tollens
Lines 2-4 indicate that the first part of the reasoning appeals to two of the premises of the argument, and it
draws a conclusion from them by modus tollens. Line 5 indicates that the reasoning goes from line 4 to 5
by double negation. Lines 6 and 7 indicate that line 5 together with the first premise lead to the sentence
on line 7 by modus ponens. Finally, lines 8 and 9 indicate that line 7 together with the third premise lead
to the conclusion of the argument.
This layout of premises and inferences constitute most of the ingredients of what we will call a derivation.
Every line consists of a line number followed by a sentence followed by a justification. The sentence on
each line either (i) occurs as a premise, and the line is justified by writing "pr", or (ii) follows from previous
lines by a rule, and the line is justified by writing the number(s) of the line(s) from which it follows, along
with a short name of the rule. The short names of the rules that we have so far are “r”, "mp", "mt", and
"dn".
Copyrighted material
Chapter One -- 13
Version of Aug 2013
CHAPTER 1 SECTION 5
The particular approach taken here is to see a derivation as carrying out a task. Each task is to show that
a sentence follows from certain things. Our derivations begin with a special line stating the task; that is,
stating what is to be shown. In the sequence of steps above it would come first, and would be of this form:
1.
Show ~W
A "show" line may be introduced at any time, and it does not need a justification, because it only states
what it is we intend to derive. All other lines need justifications.
Suppose a derivation is to be constructed, guided by the reasoning given above. Following the 'show' line
we repeat two of the premises, justifying them with the notation 'pr':
1.
2.
3.
Show ~W
~T
~P→T
pr
pr
From these two lines we infer the third by modus ponens:
1.
2.
3.
4.
Show ~W
~T
~P→T
~~P
pr
pr
2 3 mt
Next, we write 'P', indicating that we are deriving it from line 4:
1.
2.
3.
4.
5.
Show ~W
~T
~P→T
~~P
P
pr
pr
2 3 mt
4 dn
Still following out the reasoning on the previous page, we repeat another premise, and then infer 'W → T'
from it together with line 5 using modus ponens:
1.
2.
3.
4.
5.
6.
7.
Show ~W
~T
~P→T
~~P
P
P→(W→T)
W→T
pr
pr
2 3 mt
4 dn
pr
5 6 mp
Again, we insert the second premise, and infer '~W' from it and line 7 by modus tollens:
1.
2.
3.
4.
5.
6.
7
8.
9.
Show ~W
~T
~P→T
~~P
P
P→(W→T)
W→T
~T
~W
pr
pr
2 3 mt
4 dn
pr
5 6 mp
pr
7 8 mt
The reasoning ends here, where we have completed the task of showing '~W'. At this point we need a
way to indicate that the task of showing '~W' has been completed. The completion of the task is indicated
by writing "dd" after that line (meaning "direct derivation"); then the "Show" on the show line is cancelled by
drawing a line through it -- because the task has been completed -- and the steps used in that process are
boxed off. Specifically, since the sentence to be shown occurs on line 9, you may write "dd" to its right,
box all lines below the show line, and cancel the "Show":
Copyrighted material
Chapter One -- 14
Version of Aug 2013
CHAPTER 1 SECTION 5
1.
Show ~W
2.
3.
4.
5.
6.
7.
8.
9.
~T
~P→T
~~P
P
P→(W→T)
W→T
~T
~W
Cancel the "Show"
pr
pr
2 3 mt
4 dn
pr
5 6 mp
pr
7 8 mt dd
Box the lines
Write "dd"
The cancellation of the show line indicates that the task was successfully completed, and the boxing
encloses the lines used in completing that task.
It is also permissible to wait and write the "dd" on a later line. Such a line contains no sentence itself; its
justification consists of the number of the line where the target sentence occurs, followed by "dd". Here is
the same derivation with the dd justification on a later line:
1.
Show ~W
2.
3.
4.
5.
6.
7.
8.
9.
10.
~T
~P→T
~~P
P
P→(W→T)
W→T
~T
~W
pr
pr
2 3 mt
4 dn
pr
5 6 mp
pr
7 8 mt
9 dd
Empty line with "dd"
It is often a matter of taste which technique to use for indicating the completion of a direct derivation.
Here is another illustration of a direct derivation, used to show this argument valid:
Q→~S
V→X
~V→S
~X
∴ ~Q
The derivation begins with a line indicating that the task is to show the conclusion of the argument:
1.
Show ~Q
The next few lines give the reasoning steps:
2.
3.
4.
5.
6.
7.
8.
9.
V→X
~X
~V
~V→S
S
~~S
Q→~S
~Q
pr
pr
2 3 mt
pr
4 5 mp
6 dn
pr
7 8 mt
On line 9 we have completed the task. So we write "dd" and box and cancel:
Copyrighted material
Chapter One -- 15
Version of Aug 2013
CHAPTER 1 SECTION 5
1.
Show ~Q
2.
3.
4.
5.
6.
7.
8.
9.
V→X
~X
~V
~V→S
S
~~S
Q→~S
~Q
pr
pr
2 3 mt
pr
4 5 mp
6 dn
pr
7 8 mt dd
(Line 7 is necessary before using modus tollens with line 8. This form of inference is indeed valid:
□→~○
○
∴ ~□
however, it is not itself an instance of modus tollens. It is instead an inference that is easily justified using
double negation along with modus tollens.)
We indent all of the lines immediately following a show line; this is a device for keeping track (by
indentation) of where the task that is initiated by the "show" is being carried out. The indentation also
reserves a space for the box that will be drawn if the derivation is successful.
In getting precise about how to construct a direct derivation, it will help to specify what previously occurring
things can be appealed to when applying a rule. We will say that a previous line is available from a given
line just in case it is an earlier line that is not an uncancelled show line and is not already in a box:
In a derivation from a set P of premises, a line is available from a given line just in
case it is a member of P or it is an earlier line that is not an uncancelled show line
and is not already in a box.
Whether a line is available or not depends on your perspective. A show line is not available from the line
immediately below it -- because it is not yet cancelled. But once it is cancelled, it is available from all lines
below the line from which it was cancelled. And a line may be available from a given line, but once it is
boxed, it is not available from any line outside the box.
A direct derivation from a set of sentences P consists of a sequence of lines (including justifications
when appropriate) that is built up, step by step, where each step is in accordance with these provisions:
•
•
•
•
•
A show line consists of the word "Show" followed by a sentence. The first step of producing a
derivation must be to introduce a show line. A show line also may be introduced at any later step.
Show lines are not given a justification.
At any step, any sentence from the set of sentences, P, may be introduced, justified with the
notation "pr".
At any step a line may be introduced if it follows by a rule from previous available lines in the
derivation; it is justified by citing the numbers of those previous lines and the name of the rule.
If a line is introduced whose sentence is the same as the sentence in the closest previous
uncancelled show line, one may, as the next step, write "dd" at the end of that line, draw a line
through the word "Show", and draw a box around all the lines below the show line, including the
current line.
(Alternatively) At any step, if any previous available line contains a sentence that is the same as
that in the closest previous uncancelled show line, one may introduce a line with no sentence on
it, justifying it by citing the number of the earlier line followed by "dd"; one then draws a line
through the word "Show", and draws a box around all the lines below that show line, including the
current line.
Copyrighted material
Chapter One -- 16
Version of Aug 2013
CHAPTER 1 SECTION 5
These instructions show how to construct a derivation in a step by step fashion. Any number of steps
results in a derivation, but not necessarily one that completes any of the tasks set out by its show lines.
For example, when constructing a derivation above, at a certain stage we had reached this far in the
process:
1.
2.
3.
4.
Show ~W
~T
~P→T
~~P
pr
pr
2 3 mt
This sequence of lines satisfies the conditions for being a derivation as defined above. But there is a
sense in which it is not yet finished. For this purpose we define a "complete" derivation:
A derivation is complete if every show line is cancelled and every line that is not a show
line is boxed.
The point of doing a derivation is often to show that a certain argument is formally valid. When a
derivation shows that an argument is formally valid, we say that it "validates" the argument:
A derivation validates an argument if and only if it is a complete derivation from the
premises of that argument, and the conclusion of the argument appears on an unboxed
cancelled show line in the derivation.
EXERCISES
1. Check through each line of the following direct derivations to determine whether it can be constructed
by means of the provisions for direct derivations given above, where the set P is taken to be the premises
of the displayed arguments. (When assessing a given line, assume that all previous lines are correct.)
Argument:
1.
2.
3.
4.
5.
6.
7.
8.
9.
Copyrighted material
P → (Q→~R)
~P → ~Q
Q
∴ ~R
Show ~R
Q
~~Q
~P → ~Q
~~P
P
P → (Q→~R)
Q→~R
~R
pr
2 dn
pr
3 4 mt
6 dn
pr
6 7 mp
2 8 mp dd
Chapter One -- 17
Version of Aug 2013
CHAPTER 1 SECTION 5
Argument:
1.
2.
3.
4.
5.
6.
7.
8.
9.
Show ~R
Q
~P→~Q
P
R→~Q
~~Q
~~R
R
~Q
Argument:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
P → (R→~Q)
~P → ~Q
Q
∴ ~R
pr
2 pr
2 3 mt
4 mp
2 dn
5 6 mt
7 dn
5 8 mp dd
~O
S → (W→~O)
O→S
W
∴ ~S
Show ~S
W
pr
~S
pr
O→S
pr
~O
3 4 mt
W
pr
~(W→~O)
5 6 mt
S → (W→~O) pr
~S
7 8 mt
9 dd
2. Construct direct derivations to validate each of the following arguments:
P
Q → ~P
R→Q
∴ ~R
W → ~(V→~Y)
X → (V→~Y)
V→Y
(V→Y) → X
∴ ~W
(W→Z) → (Z→W)
(Z→W) → ~X
P→X
~~P
∴ ~(W→Z)
Copyrighted material
Chapter One -- 18
Version of Aug 2013
CHAPTER 1 SECTION 6
6 CONDITIONAL DERIVATIONS
In this section we learn about one of the most powerful and useful procedures for constructing proofs in a
natural way. The procedure is called conditional derivation. It is meant to reflect a natural reasoning
process that involves hypothetical inference. Suppose that you wish to show that the following argument
is valid:
If Robert drives, Sam won't drive.
If Sam doesn't drive, Teresa won't go.
Willa will go only if Teresa does.
∴ If Robert drives, Willa won't go.
If we try to reason as above using our rules mp, mt, and dn, we will not succeed; none of them apply to the
premises we are given. What you would probably do on your own is to reason somewhat as follows:
ASSUME that Robert drives
Well, if he drives, Sam won't (given); so Sam won't drive
But if Sam doesn't drive, Teresa won't go (given), so Teresa won't go.
But Willa will go only if Teresa does (given), so Willa won't go.
So, SUMMING UP, if Robert drives, Willa won't go.
The middle three steps look familiar; they are inferences from premises and previously stated sentences,
and they are all justifiable by rules that we have. But the first and last steps are new. What does it mean
to "assume", as we have done in the first step, and what is this "summing up" in the last step? What role
do these have as legitimate parts of a piece of reasoning?
Here is what goes on in "conditional" reasoning. Our goal is to show that a certain conditional sentence
follows from certain premises. (In the example above, the conditional sentence is 'If Robert drives, Willa
won't go'.) We then "assume" the antecedent of the conditional. If we can use this to derive the
consequent of the conditional, we conclude that this reasoning has shown the conditional itself to follow
from the given premises. An example:
R→~S
~S→~T
W→T
∴ R→~W
A derivation using the conditional derivation technique begins with a line specifying the task, which is to
show the conclusion. This is followed by an assumption of the antecedent of the conditional to be shown:
1.
2.
Show R→~W
R
ass cd
assumption for conditional derivation
(the goal is now to derive the consequent: ~W)
The reasoning in the center of the derivation proceeds normally:
3.
4.
5.
6.
7.
8.
R→~S
~S
~S→~T
~T
W→T
~W
pr
2 3 mp
pr
4 5 mp
pr
6 7 mt
Now that we have derived the consequent of the conditional on line 8, We cite "cd", and we box and
cancel:
Copyrighted material
Chapter One -- 19
Version of Aug 2013
CHAPTER 1 SECTION 6
1.
Show R→~W
2.
3.
4.
5.
6.
7.
8.
R
R→~S
~S
~S→~T
~T
W→T
~W
ass cd
pr
2 3 mp
pr
4 5 mp
pr
6 7 mt cd
Cancel the "Show"
Box the lines
Write "cd"
Lines 2-8 show that given the premises, we may derive '~W' from 'R'. Our conditional derivation technique
says that this amounts to showing that those premises validate 'R→~W', so we may box and cancel.
Here is another example used to show that the following very short argument is valid:
∴
S
(S→R)→R
1.
Show (S→R)→R
2.
3.
4.
S→R
S
R
ass cd
pr
2 3 mp
<the goal is now to derive the consequent: R>
We have assumed the antecedent of the conditional on the show line, and we have now succeeded in
deriving the consequent of that conditional. So we may use the technique of conditional derivation:
1.
Show (S→R)→R
2.
3.
4.
S→R
S
R
ass cd
pr
2 3 mp cd
EXERCISES
1. For each of the following derivations, determine which lines are correct and which incorrect. (In
assessing a line, assume that previous lines are correct.)
a.
P → (~Q→R)
~R
∴ P→Q
1. Show P → Q
2.
3.
4.
5.
6.
7.
b.
P
P → (~Q→R)
~Q → R
~R
~~Q
Q
ass cd
pr
2 3 mp
pr
4 5 mt
6 dn cd
P → (Q→~R)
R
∴ P → ~Q
1. Show P → ~Q
2.
3.
4.
5.
6.
7.
P
~Q
P → (Q→~R)
Q → ~R
R
~Q
Copyrighted material
ass cd
1 2 mp
pr
2 4 mp
pr
5 6 mt cd
Chapter One -- 20
Version of Aug 2013
CHAPTER 1 SECTION 6
c.
~S → (Q→R)
R → ~(Q→R)
~P → R
∴ P → ~S
1. Show P → ~S
2.
3.
4.
5.
6.
7.
8.
9.
~S
Q→R
~~(Q→R)
R → ~(Q→R)
~R
~P → R
~~P
P
ass cd
2 mp
3 dn
pr
4 5 mt
pr
6 7 mt
8 dn cd
2. Construct correct derivations for each of the following arguments using conditional derivations.
a.
P → (Q → (R→S))
~Q → ~R
R
∴ P→S
b.
Q → ~(R→S)
P → (R→S)
~Q → R
∴ P→S
c.
U → (U→V)
~R → ~(U→V)
R → ~S
∴ U → ~S
3. Symbolize the following arguments using the sentence letters given, and then give derivations to
validate them.
a.
If Seymour likes papayas he'll have them for tea. He won't have them for tea if we don't have any.
If we didn't shop yesterday, we don't have any papayas. So if Seymour likes papayas, we
shopped yesterday. (P: Seymour likes papayas; T: Seymour will have papayas for tea; X: We
have some papayas; S: We shopped yesterday.)
b.
If today is Thursday, then Saturday is two days from now. If the party is on Saturday, then if
Saturday is two days from now, then so is the party. I can't go to the party if it's two days from
now. The party is on Saturday. So if today is Thursday, I can't go to the party. (T: Today is
Thursday; S: Saturday is two days from now; P: The party is two days from now; Y: The party is
on Saturday; X: I can go to the party.)
c.
If Samantha is at home, she won't cause any trouble. If she isn't at home, she can't be reached
by telephone. If she can't be reached by telephone, it's too late to tell her about the party. If she
comes to the party she will cause some trouble. So if it's not too late to tell her about the party,
she won't come. (S: Samantha is at home' T: Samantha will cause trouble; R: Samantha can be
reached by telephone; X: It's too late to tell Samantha about the party; Y: Samantha will come to
the party.)
Copyrighted material
Chapter One -- 21
Version of Aug 2013
CHAPTER 1 SECTION 7
7 INDIRECT DERIVATIONS
There is a third technique for doing derivations, called indirect derivation. It is often used in cases where
direct derivation and conditional derivation do not obviously apply. An example is an attempt to show that
this inference is valid:
Polk was a president
Whitney wasn't a president
∴ It is not the case that if Polk was a president, so was Whitney
The conclusion is not a conditional, so a straightforward application of conditional derivation does not
seem possible. Nor is it clear how to derive the conclusion using mp, mt, or dn. To validate this argument
you might reason as follows:
We want to show that it is not the case that if Polk was a president, so was Whitney. Well,
assume the opposite: assume that it is the case that if Polk was a president then so was Whitney.
Then, since we are given that Polk was a president, so was Whitney. But we are given that
Whitney wasn't. So we are led to absurd conclusions: Whitney was a president and Whitney was
not a president. So the assumption we made, which lead to these inferences, must not be true
(given the premises of the argument).
The reasoning is called indirect because in order to show something, you assume the opposite and derive
contradictory sentences from it (along with the premises). (Sentences are contradictory when one is the
negation of the other.) If you succeed in doing this, you have shown that the negation of what you are
trying to derive isn't true; it can’t be true because it entails contradictory sentences. So what you are trying
to derive must itself be true.
The technique of indirect derivation has two parts. First, there is a new kind of assumption: immediately
following a show line you may assume the opposite of the sentence on the show line. (The opposite of the
sentence is its negation, or its "unnegation" if it is already a negation.) Then when you have two
sentences, one of which is the negation of the other, after the last one derived you add the line number of
the other and write "id" for "indirect derivation"; then you box and cancel. (Alternatively, you may write a
later line with no sentence on it, citing the line numbers of both of the contradictory sentences, write "id';
and then box and cancel.)
Here is a formal derivation corresponding to the above reasoning:
P
~W
∴ ~(P→W)
1.
2.
3.
4.
5.
Show ~(P→W)
P→W
P
W
~W
ass id
assumption for indirect derivation
pr
2 3 mp
pr
Having reached line 5 we may add to it the line number 4 and "id"; then box and cancel:
1. Show ~(P→W)
2.
3.
4.
5.
P→W
P
W
~W
Copyrighted material
ass id
pr
2 3 mp
pr 4
id
Cancel the "Show"
Box the lines
Write "id"
Chapter One -- 22
Version of Aug 2013
CHAPTER 1 SECTION 7
The "4 id" at the end of line 5 indicates that the sentence on line 4 contradicts that on the current line, line
5, and thus the assumption (on line 2) which lead to them must be false (given the premises of the
argument).
Here is another example of a derivation for this short valid argument:
U→S
~U→S
∴ S
1.
Show S
2.
3.
4.
5.
6.
~S
U→S
~U
~U→S
S
ass id
pr
2 3 mt
pr
4 5 mp
Since we have ~S on line 2 and S on line 6, we are in a position to box and cancel:
1.
Show S
2.
3.
4.
5.
6.
~S
U→S
~U
~U→S
S
ass id
pr
2 3 mt
pr
4 5 mp 2 id
Notice that we could have postponed the id step to a later line:
1.
Show S
2.
3.
4.
5.
6.
7.
~S
U→S
~U
~U→S
S
ass id
pr
2 3 mt
pr
4 5 mp
26
id
THINKING UP DERIVATIONS: Suppose that you have an argument and you are not sure whether or not
it is valid. You can show it not to be valid by producing what is usually called a "counter-example" -- a
logically possible situation in which its premises are all true and its conclusion false. If you get a counterexample, the argument is invalid. Suppose that you aren't able to find a counter-example. That might be
because the argument is valid, or it might be because you haven't been lucky enough or clever enough to
find a counter-example. So failing to find a counter-example, by itself, shows nothing. However,
sometimes when you try to find a counter-example, and you fail, this is because any attempt to make the
premises true and conclusion false leads you to assign opposite truth values to some sentence. If this
happens to you, your failure can be used as a guide to producing a derivation that validates the argument.
Typically, it is a guide to producing an indirect derivation.
Here is an example. You are given the argument:
Copyrighted material
Chapter One -- 23
Version of Aug 2013
CHAPTER 1 SECTION 7
If Pedro left early then if Taylor stayed, Zack left
Taylor stayed
If Zack left, Pedro didn't leave early
∴ Pedro didn't leave early
Symbolized, it is:
P → (T → Z)
T
Z → ~P
∴ ~P
Consider an attempt to produce a counter-example. You need a case in which the conclusion, '~P', is
false; that is, one in which 'P' is true. So assume that 'P' is true. By the first premise, that makes 'T→Z'
be true, and then by the second premise, 'Z' must be true. Then by the third premise, '~P' must be true.
Oops! We can't make both 'P' and '~P' true. So there must be no counter-example.
Here is a derivation based on that reasoning. The assumption that 'P' is true is just like an assumption for
purposes of indirect derivation:
1. Show ~P
2.
P
ass id
Now the next few steps of our attempt to get a counter-example are paralleled by familiar steps in a
derivation:
1. Show ~P
2.
3.
4.
5.
6.
7.
8.
P
ass id
P → (T→Z) pr
T→Z
2 3 mp
T
pr
Z
4 5 mp
Z → ~P
pr
~P
6 7 mp
The "Oops" in the failed counter-example search parallels the fact that we now have contradictories on
lines 2 and 8. So we may box and cancel:
1. Show ~P
2.
3.
4.
5.
6.
7.
8.
P
P → (T→Z)
T→Z
T
Z
Z → ~P
~P
ass id
pr
2 3 mp
pr
4 5 mp
pr
6 7 mp 2 id
This then is an indirect derivation that validates the argument.
Copyrighted material
Chapter One -- 24
Version of Aug 2013
CHAPTER 1 SECTION 7
EXERCISES
1. For each of the following derivations, determine which lines are correct and which incorrect. (In
assessing a line, assume that previous lines are correct.)
a.
R → (S → T)
S
~T
∴ ~R
1. Show ~R
2.
3.
4.
5.
6.
7.
b.
R
R → (S → T)
S→T
S
T
~T
ass id
pr
2 3 mp
pr
4 5 mp
pr 6 id
~S → P
R
S → ~R
∴ P
1. Show P
2.
3.
4.
5.
6.
7.
c.
R
~S → ~R
~S
~P
~S → P
~~S
pr
pr
2 3 mt
ass id
pr
5 6 mt 5 id
U → (V→~W)
X → (U→V)
(V → ~W) → X
∴ U→V
1. Show U → V
2.
3.
4.
5.
6.
7.
U → ~V
U → (V→~W)
~V
X → (U→V)
~X
~(U→V)
Copyrighted material
ass id
pr
2 3 mt
pr
2 5 mt
5 6 mt 2 id
Chapter One -- 25
Version of Aug 2013
CHAPTER 1 SECTION 7
2. Construct correct derivations for each of the following arguments using indirect derivations.
a.
~Q → R
S → ~R
~S → Q
∴ Q
b.
(P→Q) → R
S → (P→Q)
~S → R
∴ R
c.
~P → (R→S)
(R→S) → T
~T
Q → (R→S)
∴ ~(P → Q)
Copyrighted material
Chapter One -- 26
Version of Aug 2013
CHAPTER 1 SECTION 8
8 SUBDERIVATIONS
Reasoning can be intricate. One way in which this happens is when derivations occur within derivations.
Consider the following argument in symbolic form:
~(R→Q) → P
P → (~Q→Q)
~Q
∴ ~R
1.
Show ~R
It is not clear how to directly derive ~R using our rules mp, mt, and dn. A conditional derivation is not
applicable because ~R is not a conditional. One could assume R for purposes of doing an indirect
derivation, but it is not clear how to proceed from there. An alternative approach is to use a derivation
within the main derivation. Here is one way to proceed:
Try to derive the negation of ‘~Q→Q’ from the third premise, and then use modus tollens on the
second, and then on the first, premise to get R→Q, which then leads to the desired conclusion
using mt and the third premise.
The first part of this strategy -- deriving ‘~(~Q→Q)’ from the third premise -- requires a derivation of its
own. You can do this as a conditional derivation, leaving you with:
1.
2.
Show ~R
Show ~(~Q→Q)
3.
4.
5.
~Q→Q
~Q
Q
ass id
pr
3 4 mp 4 id
You now may use line 2 just as you would use any other line; once the "Show" is cancelled, the line is no
longer something you are trying to derive, it is something you have derived. (A cancelled 'show' means
"shown".) You proceed:
1.
2.
Show ~R
Show ~(~Q→Q)
3.
4.
5.
~Q→Q
~Q
Q
ass id
pr
3 4 mp 4 id
6.
7.
8.
9.
10.
11.
12.
P → (~Q→Q)
~P
~(R→Q) → P
~~(R→Q) 7 8
R→Q
~Q
~R
pr
2 6 mt
pr
mt
9 dn
pr
10 11 mt
This is a complete direct derivation, so you may box and cancel:
Copyrighted material
Chapter One -- 27
Version of Aug 2013
CHAPTER 1 SECTION 8
1.
2.
Show ~R
Show ~(~Q→Q)
3.
4.
5.
~Q→Q
~Q
Q
ass id
pr
3 4 mp 4 id
6.
7.
8.
9.
10.
11.
12.
P → (~Q→Q)
~P
~(R→Q) → P
~~(R→Q)
R→Q
~Q
~R
pr
2 6 mt
pr
7 8 mt
9 dn
pr
10 11 mt dd
The subderivation was just a way to derive '~(~Q→Q)'.
Here is another example:
(P→~Q) → (R →S)
Q→~P
T→R
∴ T→S
We begin the derivation by stating what is to be shown:
1.
Show T→S
Generally, the easiest way to derive a conditional is to use conditional derivation. So we write this:
1. Show T→S
2.
T
ass cd
with the goal of deriving S, thereby completing the conditional derivation. At this point it is unclear what to
do next. However, we notice that S occurs only once in the premises -- in the first premise. And if we
could derive the antecedent of that premise, we could use mp to infer R→S, which would get us close to
completing the derivation. So we try to derive the antecedent of the first premise:
1.
2.
3.
Show T→S
T
Show P→~Q
ass cd
Line 3 is a conditional, so we try to use a conditional derivation:
1. Show T→S
2.
3.
T
Show P→~Q
4.
P
ass cd
ass cd
Our immediate goal is now to derive ~Q. We can do that by appealing to the double negation of P along
with the second premise:
1.
Show T→S
2.
3.
T
Show P→~Q
ass cd
4.
5.
6.
7.
P
~~P
Q→~P
~Q
ass cd
4 dn
pr
5 6 mt
Copyrighted material
Chapter One -- 28
Version of Aug 2013
CHAPTER 1 SECTION 8
This completes the conditional subderivation, so we box and cancel:
1.
Show T→S
2.
3.
T
Show P→~Q
ass cd
4.
5.
6.
7.
P
~~P
Q→~P
~Q
ass cd
4 dn
pr
5 6 mt
We may now do the rest of the derivation:
1.
Show T→S
2.
3.
T
Show P→~Q
ass cd
4.
5.
6.
7.
P
~~P
Q→~P
~Q
ass cd
4 dn
pr
5 6 mt cd
8.
9.
10.
11.
12.
(P→~Q) → (R→S) pr
R→S
3 8 mp
T→R
pr
R
2 10 mp
S
9 11 mp
Line 12 completes the main conditional derivation, so we box and cancel, and we are done:
1.
Show T→S
2.
3.
T
Show P→~Q
ass cd
4.
5.
6.
7.
P
~~P
Q→~P
~Q
ass cd
4 dn
pr
5 6 mt cd
8.
9.
10.
11.
12.
(P→~Q) → (R→S) pr
R→S
3 8 mp
T→R
pr
R
2 10 mp
S
9 11 mp cd
Now that we have derivations within derivations, the availability of previous lines for the purpose of
applying rules can change from line to line; a line that is not available at one point can become available
later, and one that is available may become unavailable. Examples:
At line 4 above, line 3 is not available, because from the point of view of line 4, line 3 is an
uncancelled show line. But the 'show' on line 3 is cancelled at line 7, so from the point of view of
line 8, line 3 is available.
On the other hand, at line 6, line 5 is available; but line 5 is no longer available at line 8, because it
has been boxed at line 7.
We can now state explicitly what may appear in a derivation which may contain subderivations:
Copyrighted material
Chapter One -- 29
Version of Aug 2013
CHAPTER 1 SECTION 8
DERIVATIONS
A derivation from a set of sentences P consists of a sequence of lines that is built up in order, step by
step, where each step is in accordance with these provisions:
•
•
•
•
•
•
•
•
Show line: A show line consists of the word "Show" followed by a symbolic sentence. A show
line may be introduced at any step. Show lines are not given a justification.
Premise: At any step, any symbolic sentence from the set P may be introduced, justified with the
notation "pr".
Rule: At any step, a line may be introduced if it follows by a rule from sentences on previous
available lines; it is justified by citing the numbers of those previous lines and the name of the rule.
Direct derivation: When a line (which is not a show line) is introduced whose sentence is the
same as the sentence on the closest previous uncancelled show line, one may, as the next step,
write "dd" following the justification for that line, draw a line through the word "Show", and draw a
box around all the lines below the show line, including the current line.
Assumption for conditional derivation: When a show line with a conditional sentence is
introduced, as the next step one may introduce an immediately following line with the antecedent
of the conditional on it; the justification is "ass cd".
Conditional derivation: When a line (which is not a show line) is introduced whose sentence is
the same as the consequent of the conditional sentence on the closest previous uncancelled
show line, one may, as the next step, write "cd" at the end of that line, draw a line through the
word "Show", and draw a box around all the lines below the show line, including the current line.
Assumption for indirect derivation: When a show line is introduced, as the next step one may
introduce an immediately following line with the [un]negation of the sentence on the show line; the
justification is "ass id".
Indirect derivation: When a sentence is introduced on a line which is not a show line, if there is a
previous available line containing the [un]negation of that sentence, and if there is no uncancelled
show line between the two sentences, as the next step you may write the line number of the first
sentence followed by "id" at the end of the line with the second sentence. Then you cancel the
closest previous "show", and box all sentences below that show line, including the current line.
Except for steps that involve boxing and canceling, every step introduces a line. When writing out a
derivation, every line that is introduced is written directly below previously introduced lines.
Optional variant: When boxing and canceling with direct or conditional derivation, the "dd" or "cd"
justification may be written on a later line which contains no sentence at all, and which is followed by the
number of the line that satisfies the conditions for direct or conditional derivation. With indirect derivation,
the "id" justification may be written on a later line which contains no sentence at all, and which is followed
by the numbers of the two lines containing contradictory sentences. In both cases, the lines cited must be
available from the later line.
Copyrighted material
Chapter One -- 30
Version of Aug 2013
CHAPTER 1 SECTION 8
EXERCISES
1. For each of the following derivations, determine which lines are correct and which incorrect. (In
assessing a line, assume that previous lines are correct.)
Tip: When a box occurs in a correctly formed derivation, it is put there by the command ('dd', or 'cd' or 'id')
that appears at the end of the last line within the box. When a "show" is cancelled, it is cancelled by the
same command that puts the box immediately below the "show'.
a.
P → (Q→R)
~Q → S
∴ P → (~S→R)
1. Show P→(~S→R)
2.
3.
P
Show ~S → R
ass cd
4.
5.
6.
7.
8.
9.
10.
~S
~Q → S
~~Q
Q
P → (Q→R)
Q→R
R
ass cd
pr
4 5 mt
6 dn
pr
2 8 mp
7 9 mp cd
11.
b.
3 cd
R→Q
Q→P
∴ R→P
1. Show R → P
2.
3.
4.
5.
6.
7.
8.
R
R→Q
Q
Show P
~P
Q→P
~Q
9.
c.
ass cd
pr
2 3 mp
ass id
pr
6 7 mt 4 id
error on this line!
5 cd
P→Q
(R→Q) → S
(U→S) → ~P
∴ ~P
Copyrighted material
Chapter One -- 31
Version of Aug 2013
CHAPTER 1 SECTION 8
1. Show ~P
2.
3.
P
Show R → Q
4.
5.
6.
R
P→Q
Q
7.
Show U→ S
ass id
ass cd
pr
2 6 mp cd
8.
9.
10.
U
(R→Q) → S
S
ass cd
pr
3 9 mp cd
11.
12.
13.
(U→S) → ~P
~P
P
pr
7 11 mp
2 r 12 id
2. Construct correct derivations for each of the following arguments.
a.
P → (Q→R)
S→Q
∴ S → (P→R)
b.
(P→Q) → Q
P→R
Q → ~Q
∴ ~(R → Q)
c.
(U → V) → (W→X)
U→Z
~V → ~Z
X→Z
∴ W→Z
Copyrighted material
Chapter One -- 32
Version of Aug 2013
CHAPTER 1 SECTION 9
9 SHORTCUTS
Writing long derivations can be tedious. Here are two shortcuts.
Citing premises without rewriting them
Every time we have wanted to appeal to a premise when using a rule of inference, we have written the
premise into the derivation, justifying it by 'pr', and then when we use the rule we cite the number of the
line where the premise was written. We can skip writing down the premise entirely if the justification of the
rule identifies the premise by name. For example, instead of having:
8. Q → R
9. Q
10. R
pr
(where 'Q' is the third premise)
8 9 mp
we just write:
8.
9.
Q→R
R
8 pr3 mp
This is equivalent to just assuming that the premises all come with line numbers: pr1, pr2, pr3, . . ., and
citing those line numbers when we use a rule of inference. (Our directions above for constructing
derivations already include this option.)
For example, here is a derivation we gave earlier:
Premises:
Conclusion:
1.
P→(W→T)
~P→T
~T
∴ ~W
Show ~W
2.
3.
4.
5.
6.
7.
8.
9.
~T
~P→T
~~P
P
P→(W→T)
W→T
~T
~W
pr
pr
2 3 mt
4 dn
pr
5 6 mp
pr
7 8 mt dd
Using the shortcut, we can essentially skip lines 2, 3, 6, 8, to get this shortened derivation which has
analogues of lines 1, 4, 5, 7, 9 in the original derivation:
1.
Show ~W
2.
3.
4.
5.
~~P
P
W→T
~W
pr2 pr3 mt
2 dn
3 pr1 mp
4 pr3 mt dd
Explicitly, the technique is:
Citing Premises in Rules
When citing a premise in applying a rule of inference, use 'pr1', 'pr2' 'pr3', . . . to identify
the first, second, third, . . . premises.
Copyrighted material
Chapter One -- 33
Version of Aug 2013
CHAPTER 1 SECTION 9
Mixed derivations: Our derivation rules are already formulated in a way that lets you use any one of dd,
cd, id, in cases in which you expect to use another of them. For example, suppose you are trying to show
'P→Q' by cd, assuming 'P'. But you derive 'P→Q' instead of 'Q', Then you can use dd to box and cancel.
The fact that you have assumed 'P' for purposes of producing a conditional derivation does not interfere
with this use of dd. Given these lines:
1. Show □→○
2. □
3. ……..
7.
8.
ass cd
……..
□→○
you can box and cancel:
1. Show □→○
2.
3.
□
ass cd
……..
7.
8.
……..
□→○
dd
This is a "mixed" derivation: an assumption is made for constructing a conditional derivation, and then 'dd'
is used instead of 'cd' to complete it. That's OK because this is just a shortcut. Whenever you are in the
position described above, you could instead add a step to the end of the derivation and then conclude the
derivation as a conditional derivation. Just add step 9:
1.
2.
3.
4.
7.
8.
9.
Show □→○
□
ass cd
……..
……..
……..
□→○
○
2 8 mp
and then box and cancel using cd:
1. Show □→○
2.
3.
4.
□
……..
……..
7.
8.
9.
……..
□→○
○
ass cd
2 8 mp cd
So using dd at the end of line 8 is merely a way to save a step.
Copyrighted material
Chapter One -- 34
Version of Aug 2013
CHAPTER 1 SECTION 9
Similarly, if you are trying to do an indirect derivation and you end up deriving the sentence on the show
line, you may use dd. That is, if you have:
1.
2.
3.
4.
Show □
~□
……..
……..
7.
8.
……..
□
ass id
you may box and cancel with dd:
1.
Show □
2.
3.
4.
~□
……..
……..
7.
8.
……..
□
ass id
dd
Here the shortcut is obvious; you are already in a position to use id since you already have the
contradictory sentences that you need; you can instead cite the line number of the other contradictory and
use id:
1.
Show □
2.
3.
4.
~□
……..
……..
7.
8.
……..
□
ass id
2 id
Similarly you can use cd when you are set up for a direct derivation of a conditional, or when you are trying
to derive a conditional using id; and you can use id when you have derived contradictories even if you are
set up for a direct or conditional derivation.
In allowing for mixed derivations, we are not actually changing anything. Our rules already allow for them.
So this is a summary of things we can already do with the rules as stated:
Mixed derivations
You may use dd, cd, and id to complete a derivation by boxing and canceling
whenever they apply, whether or not an assumption has been made, and
regardless of the type of assumption if any.
Copyrighted material
Chapter One -- 35
Version of Aug 2013
CHAPTER 1 SECTION 9
EXERCISES
1. Each of the following derivations is a mixed derivation. In each case produce another derivation which
is not mixed.
a.
P→R
Q → ~R
~Q → Q
∴ P→Q
1. Show P → Q
2.
3.
4.
5.
6.
7.
8.
9.
b.
ass cd
pr
2 3 mp
4 dn
pr
5 6 mt
pr
7 8 mp 7 id
Q→U
Q → ~U
R→Q
R
∴ P
1. Show P
2.
3.
4.
5.
6.
7.
8.
c.
P
P→R
R
~~R
Q → ~R
~Q
~Q → Q
Q
R
R→Q
Q
Q→U
U
Q → ~U
~U
pr
pr
2 3 mp
pr
4 5 mp
pr
4 7 mp 6 id
U → (V→W)
X→U
~X → W
∴ V→W
1. Show V → W
2.
3.
4.
5.
6.
7.
8.
~(V → W)
U → (V→W)
~U
X→U
~X
~X → W
W
Copyrighted material
ass id
pr
2 3 mt
pr
4 5 mt
pr
6 7 mp cd
Chapter One -- 36
Version of Aug 2013
CHAPTER 1 SECTION 9
2. Do the derivations from 1a-c above using premise line numbers instead of rule pr.
3. Earlier we stipulated that a conditional sentence is false when its antecedent is true and its consequent
false, and true in all other cases. This was not arbitrary. Given the rules and derivation procedures that
we have adopted, these choices are forced on us. For, using our rules and procedures, we can produce
derivations to show each of the following arguments to be valid:
P
Q
∴ P→Q
P
~Q
∴ ~(P→Q)
~P
Q
∴ P→Q
~P
~Q
∴ P→Q
The first of these tells us that when the antecedent and consequent of a conditional are both true, so is the
conditional. The second tells us that when the antecedent of a conditional is true and the consequent
false, the conditional is false. The third and fourth tell us that in either case in which the antecedent is
false, the conditional is true.
Produce short derivations for each of those four arguments.
Copyrighted material
Chapter One -- 37
Version of Aug 2013
CHAPTER 1 SECTION 10
10 STRATEGY HINTS FOR DERIVATIONS
Some strategies are generally useful in thinking up how to construct derivations. They do not always
work, but they are often the best way to begin the search for a successful derivation.
1. Try to reason out the argument for yourself. If you can do that, then write down the steps you went
through in your own reasoning. These steps will often be an outline of a good derivation. This idea of
reasoning things out and then turning the steps into a derivation has been illustrated above when
introducing direct, conditional, and indirect derivations. This is by far the best approach to thinking up a
derivation.
2. Begin with a sketch of an outline of a derivation, and then fill in the details. For example, in
thinking up a derivation for this argument:
P → (Q→R)
Q
∴ P→R
you might begin by saying: "I'll do a conditional derivation: I'll assume P and then use this together with
the premises to derive R. Then I'll box and cancel by cd." This outline gives you this much of a derivation:
1.
Show P→ R
2.
P
ass cd
:::::
:::::
:::::
13.
R
cd
. . . . then box and cancel
All you need to do then is to fill in lines 3 to 12. (I'm just guessing that it will take exactly eight more steps
to finish the derivation. If it takes more or less, change the '13' to the appropriate number.)
3. Write down obvious consequences: Write down obvious consequences of premises or of sentences
that have already been derived. If you write down the simple consequences of things that you already
have, then you have lots of resources right in front of you. Of course, you can do this in your head instead
of on paper, and sometimes that is sufficient. But if things are not completely clear to you, writing down
obvious consequences can be useful.
Suppose you are trying to construct a derivation to validate this argument:
P → (Q→R)
S→Q
Q→P
S
∴ R
Let's say that you have written down the show line:
1. Show R
but you are momentarily stuck. Write down some obvious consequences of the premises:
2.
3.
4.
Q
P
Q→R
Copyrighted material
pr4 pr2 mp
pr3 2 mp
pr1 3 mp
Chapter One -- 38
Version of Aug 2013
CHAPTER 1 SECTION 10
At this point, if you look over what is available, it will be obvious that you can get the desired conclusion in
one additional step:
5.
R
2 4 mp
4. If you are trying to derive a conditional, use conditional derivation. This is almost always the
easiest way to derive a conditional. This has been amply illustrated above.
5. If you have a conditional, try to derive its antecedent (and then use modus ponens), or try to
derive the negation of its consequent (and then use modus tollens). Here is an illustration:
P → (Q → R)
S→Q
~P → ~S
S
∴ R
The only premise in which R occurs is the first one, which is a conditional. You could apply modus ponens
to this conditional if you could prove P. Our strategy rule suggests that you try to derive P. That is not
difficult to do:
1. Show R
2. ~~S
3. ~~P
4. P
pr4 dn
2 pr3 mt
3 dn
This immediately gives us:
5.
Q→R
4 pr1 mp
It is then easy to complete the derivation:
6.
7.
Q
R
pr2 pr4 mp
5 6 mp
and you are ready to box and cancel.
6. Try indirect derivation. When you reach a place where none of the other strategies clearly apply,
assume the negation of what you are trying to derive and try to derive a contradiction. This too has been
amply illustrated above.
7. When doing an indirect derivation, try to derive the negation of a premise or the negation of
something that you already have derived. This is especially useful if you already have the negation of
a conditional, for you can try to derive the conditional by using conditional derivation, and then you will
have both the conditional and its negation, and you can box and cancel with id. An example. You have:
R
(Q → S) → ~R
∴ ~S
Now begin a derivation, and follow strategy rule 1: write out obvious consequences of what you have:
1. Show ~S
2. S
3. ~~R
4. ~(Q→S)
ass id
pr1 dn
3 pr2 mt
The moves so far are pretty straightforward. But it may not be clear what to do next. Strategy rule 7
suggests that you try to derive the conditional 'Q→S', in order to contradict '~(Q→S)'. You do this with a
conditional subderivation:
Copyrighted material
Chapter One -- 39
Version of Aug 2013
CHAPTER 1 SECTION 10
5.
Show Q → S
6.
7.
Q
S
ass cd
2 r cd
(this step is obvious once you notice it)
Having derived the conditional on line 5 and its negation on line 4 the indirect derivation is just about
complete. The complete derivation is:
1. Show ~S
2.
3.
4.
5.
S
~~R
~(Q→S)
Show Q → S
6.
7.
Q
S
8.
ass id
pr1 dn
3 pr2 mt
ass cd
2 r cd
4 5 id
EXERCISES
Produce correct derivations to validate these arguments.
1.
S
(R→S) → W
∴ W
2.
P → (S→R)
P → (W→S)
W→P
∴ W→R
3.
(P→Q) → S
S→T
~T → Q
∴ T
Copyrighted material
Chapter One -- 40
Version of Aug 2013
CHAPTER 1 SECTION 11
11 THEOREMS
A truth of logic (a sentence that is logically true) is a sentence that is true in any logically possible situation.
It must be true no matter what. Because of this, a truth of logic is like the conclusion of a valid argument
which has no premises. If such an argument is valid, it does not have all true premises and a false
conclusion in any logically possible situation. When there aren’t any premises, this is equivalent to saying
that it does not have a false conclusion in any logically possible situation. That is, its conclusion is true in
every logically possible situation. It is a truth of logic.
Since derivations show arguments valid, if a derivation is used to show an argument with no premises to
be valid, that amounts to showing that the conclusion is logically true. There is a special word, ‘theorem’,
that refers to any sentence shown by a technique like a derivation when no premises at all are used. So
the topic of this section is Theorems.
It is customary to indicate a theorem by placing a "therefore" sign in front of it, as if it were an argument
with its premises missing. So writing "∴□" indicates that □ is a theorem.
Here is a derivation to show that ‘P→P’ is a theorem:
∴ P→P
1.
Show P→P
2.
3.
P
ass cd
2 cd
In simple cases, theorems are obviously trivial statements. Even when they are complex, they are still
trivial in the sense that they say nothing beyond what is logically true. In this book we will list several
theorems, giving them the names "T1", "T2", and so on. Here are some given in increasing degrees of
complexity. Some of them have common names; these are indicated to the right.
T1
P→P
<just proved>
T2
Q → (P→Q)
1.
Show Q → (P→Q)
2.
3.
Q
Show P→Q
4.
5.
P
Q
6.
T3
ass cd
ass cd
2 r cd
3 cd
P → ((P→Q) → Q)
1.
2.
3.
Show P → ((P→Q) → Q)
P
ass cd
Show (P→Q) → Q
4.
5.
6.
P→Q
Q
ass cd
2 4 mp cd
3 cd
T4
(P→Q) → ((Q→R) → (P→R))
"Syllogism"
T5
(Q→R) → ((P→Q) → (P→R))
"Syllogism"
Copyrighted material
Chapter One -- 41
Version of Aug 2013
CHAPTER 1 SECTION 11
T6
(P → (Q→R)) → ((P→Q) → (P→R))
1.
"Distribution of → over →"
Show (P → (Q→R)) → ((P→Q) → (P→R))
2.
3.
P → (Q→R)
Show (P→Q) → (P→R)
4.
5.
P→Q
Show P→R
6.
7.
8.
9.
P
Q
Q→R
R
ass cd
ass cd
ass cd
4 6 mp
2 6 mp
7 8 mp cd
10.
5 cd
11.
3 cd
T7
((P→Q) → (P→R)) → (P→(Q → R))
"Distribution of → over →"
T8
(P→ (Q→R)) → (Q → (P→R))
Commutation
T9
(P → (P→Q)) → (P→Q)
T10
((P→Q) → Q) → ((Q→P) → P)
T11
~~P → P
Double negation
T12
P → ~~P
Double negation
T13
(P→Q) → (~Q → ~P)
Transposition
T14
(P → ~Q) → (Q → ~P)
Transposition
T15
(~P → Q) → (~Q → P)
Transposition
T16
(~P → ~Q) → (Q → P)
Transposition
T17
P → (~P→Q)
T18
~P → (P→Q)
T19
(~P→P) → P
Reductio ad absurdum
T20
(P→~P) → ~P
Reductio ad absurdum
T21
~(P→Q) → P
T22
~(P→Q) → ~Q
T23
((P→Q) → P) → P
Peirce's law
EXERCISES
1. Prove theorems T8, T11, T13, T17, T19, T21, T22.
Copyrighted material
Chapter One -- 42
Version of Aug 2013
CHAPTER 1 SECTION 12
12 USING PREVIOUSLY PROVED THEOREMS IN DERIVATIONS
One major use of theorems is to avoid repeating derivations that you have done earlier. For example,
suppose you have previously derived 'P → (P → (Q→P))' from no premises. Now you are doing a new
derivation where you have derived 'R', and you want to derive 'S→R'. By following out the reasoning from
earlier, but using 'R' instead of 'P', you could derive this: 'R → (R → (S→R))', and then use modus ponens
twice to get 'S → R'. But why write out the new derivation of 'R → (R → (S→R))' when it may be lengthy,
and it just involves repeating moves that you made earlier? It would be nice to have a way to "reuse" old
derivations for new purposes when the reasoning is the same. One way to do this is with this new
procedure that we adopt hereafter:
Theorems: Any instance of any previously derived theorem may be entered on any
line of a derivation. As justification, write the name of the theorem. (E.g. ‘T13’.)
An instance of a theorem is what you get by considering the theorem as a pattern, and filling in the pattern
uniformly with sentences. For example, T1 is ∴P→P. This validates the pattern ‘□→□’. Anything got by
filling in that pattern, putting the same sentence in for each occurrence of □, can be written on any line of
a derivation. For example, you can write:
(S→W) → (S→W)
21.
T1
More complicated theorems are more useful. For example, T4 is:
(P→Q) → ((Q→R) → (P→R))
This gives us the pattern:
(□→○) → ((○→△) → (□→△))
Putting ~R in for □, (U→V) for ○, and W for △ we have:
(~R→(U→V)) → (((U→V)→W) → (~R→W))
Here is an example of a use of this theorem. Suppose you wish to do a derivation for this argument:
R→~S
~S→~T
∴ R→~T
A very short derivation could be given using theorem 4:
1.
2.
3.
4.
Show R→~T
(R→~S) → ((~S→~T) → (R→~T))
(~S→~T) → (R→~T)
R→~T
Copyrighted material
T4
2 pr1 mp
3 pr2 mp dd
Chapter One -- 43
Version of Aug 2013
CHAPTER 1 SECTION 12
EXERCISES
1. Produce short derivations for these arguments using instances of the theorems listed above.
a.
X → ~(Y → Z)
∴ (Y → Z) → ~X
b.
R → (~P→S)
R → ~P
∴ R→S
c.
~(R → (S→T))
R→P
P → (Q → (S→T))
∴ ~Q
d.
Q→R
R→S
∴ Q→S
Copyrighted material
Chapter One -- 44
Version of Aug 2013
Answers to the Exercises -- CHAPTER 1
Answers to the Exercises -- Chapter 1
SECTION 1
1. a
Sentence, official notation
~~~P
|
~~P
|
~P
|
P
b
Sentence, informal notation
~Q→~R
/\
~Q ~R
|
|
Q
R
c
d
e
Not a sentence; it is impossible to construct "~→"
Not a sentence; the rules of formation do not allow you to enclose a negated formula in
parentheses, only conditionals.
Sentence, informal notation
(P→Q) → (R→~Q)
/\
(P→Q) (R→~Q)
/\
/\
P Q
R ~Q
|
Q
f
g
h
Not a sentence; conditionals contained inside other conditionals must be surrounded by
parentheses.
Not a sentence; this is the same as sentence (f) with extra parentheses put on the outside; If (g)
were a sentence, then (f) would be a sentence in informal notation; since (f) is not a sentence, (g)
can't be either.
Sentence; informal notation
(~S→R) → ((~R→S) → ~(~S→R))
/\
~S→R
(~R→S) → ~(~S→R)
/\
/\
~S R
~R→S
~(~S→R)
|
/\
|
S
~R S
~S→R
|
/\
R
~S R
|
S
Copyrighted material
Chapter One -- 45 -- Answers to the Exercises
Version of Aug 2013
Answers to the Exercises -- CHAPTER 1
i
Sentence; informal notation
P → (Q→P)
/\
P Q→P
/\
Q P
SECTION 2
1. a
b
c
d
e
2. a
S→R, because "only if" immediately precedes the consequent
R→S, because "provided that" is equivalent to "if", and "if" immediately precedes the antecedent
~S, because "won't" means the same as "will not", the sentence only contains one negation
indicator
S→R, because "only if" immediately precedes the consequent
R→S, because "given that" is equivalent to "if" and "if" immediately precedes the antecedent
Susan will be late only provided that it rains
Susan will be late only if it rains
S→R
b
Only on condition that it rains will Susan be late
Only if it rains will Susan be late
S→R
c
Susan will be late only in case it rains
Susan will be late only if it rains
S→R
d
Susan will be late only if it rains
S→R
e
It is not the case that Susan will be late
~S
SECTION 3
1. a
If Veronica doesn’t leave William won’t either
If Veronica doesn’t leave William won’t leave
If Veronica doesn’t leave then William won’t leave
~V→~W
b
William will leave if Yolanda does, provided that Veronica doesn’t
William will leave if Yolanda does, if Veronica doesn’t [leave]
If Veronica doesn't [leave], then (William will leave if Yolanda does)
If Veronica doesn't [leave], then (if Yolanda [leaves], then William will leave)
~V → (Y→W)
c
If Yolanda doesn’t leave, then Veronica will leave only if William doesn’t
~Y → (V → ~W)
d
If Yolanda doesn’t leave then Veronica will leave, given that William doesn’t
If Yolanda doesn’t leave then Veronica will leave, if William doesn’t
Copyrighted material
Chapter One -- 46 -- Answers to the Exercises
Version of Aug 2013
Answers to the Exercises -- CHAPTER 1
If William doesn’t [leave], then (if Yolanda doesn’t leave then Veronica will leave)
~W → (~Y→V)
2. a
William will leave if Veronica does
If Veronica [leaves], then William will leave
V→W
b
Veronica won't leave if William does
If William [leaves], then Veronica won't leave
W → ~V
c
If Veronica leaves, then if William doesn't leave, Yolanda will leave
If Veronica leaves, then (~W → Y)
V → (~W → Y)
d
If Veronica doesn't leave if William doesn't, then Yolanda won't
If (if William doesn't [leave], then Veronica doesn't leave), then Yolanda won't [leave]
(~W → ~V) → ~Y
e
William won’t leave provided that Veronica doesn’t leave
William won’t leave if Veronica doesn’t leave
If Veronica doesn't leave, then William won't leave
~V → ~W
f
If William leaves, then if Veronica leaves so will Yolanda
If William leaves, then (if Veronica leaves, then Yolanda will leave)
W → (V → Y)
g
William will leave only if if Veronica leaves then so will Yolanda
William will leave only if (if Veronica leaves then so will Yolanda)
William will leave → (if Veronica leaves then so will Yolanda)
William will leave → (if Veronica leaves then Yolanda will leave)
W → (V → Y)
h
William will leave only if Veronica leaves, only provided that Yolanda will leave
(William will leave only if Veronica leaves) only if Yolanda will leave
(W → V) → Y
SECTION 4
1. a
b
c
d
e
f
g
h
i
None of the above; it might look like a modus tollens inference, but the second premise, Q, would
have to first be changed to ~~Q by applying double negation; so while the argument is valid, it is
not a one-step application of modus tollens.
None of the above
Double negation
Modus ponens
Modus tollens
None of the above
Modus ponens; the consequent of the conditional does not need to be an atomic sentence, it can
be molecular as well.
None of the above; it may look like a modus tollens inference, but the second premise is not
actually the negation of the consequent of the first premise.
Double negation and none of the above are both good answers; the conclusion can be inferred by
double negation from the first premise, but since the second premise is not involved in that
Copyrighted material
Chapter One -- 47 -- Answers to the Exercises
Version of Aug 2013
Answers to the Exercises -- CHAPTER 1
inference, the whole argument is not a double-negation inference
2.
a
b
c
d
e
f
g
h
i
In all cases we can validly infer any sentence which results from putting '~~' in front of either
premise by the rule double negation. Such results are not enumerated below.
~X may be inferred by modus ponens.
X may be inferred by double negation; ~~W may be inferred by modus tollens.
Nothing additional
(R→X) may be inferred by modus ponens.
Nothing additional; Modus tollens cannot be applied because the second premise is not actually
the negation of the consequent of the first premise.
(W→X) may be inferred from the first premise by double negation; this must be done before you
apply modus tollens with the second premise, so you can't apply modus tollens in one step to get
~W.
Nothing additional; if you apply double negation to the second premise you can then apply modus
tollens as a second step to get ~W.
Nothing additional
(W→X) follows by double negation.
SECTION 5
1. Only errors are listed.
First derivation
Line 6 -- line 6 is not available at line 6; derivation can be corrected by writing "5 dn".
Second derivation
Line 3 -- when justifying writing a premise, no line citation is given
Line 4 -- the sentence on line 2 is not the negation of the consequent of the sentence on line 3; in this
case we would need to first apply dn to line 2 as an intermediate step. Then we could apply mt
with line 3, which would result in ~~P. P could then be inferred on the following line by dn.
Line 5 -- two lines must be cited with mp; the sentence inferred does follow from line 4 together with the
first premise, but the first premise must be cited somehow.
Line 7 -- "5 6 mt" would result in ~R rather than ~~R.
Line 9 -- "5 8 mp" is OK, but the derivation is not done; we set out to show ~R, but line 9 displays ~Q, so
we cannot conclude the derivation at this point and so it is incorrect to write "dd" to mark the
conclusion of the derivation of ~R.
Third derivation
Line 3 -- "~S" is not one of the premises.
Line 7 -- neither line 5 nor line 6 is a conditional, so mt cannot possibly apply to that pair of lines.
2. In each case the derivation displayed does not represent the only possible derivation; many alternate,
equally correct derivations can be given.
P
Q → ~P
R→Q
∴ ~R
Copyrighted material
Chapter One -- 48 -- Answers to the Exercises
Version of Aug 2013
Answers to the Exercises -- CHAPTER 1
1.
2.
3.
4.
5.
6.
7.
8.
Show ~R
P
~~P
Q → ~P
~Q
R→Q
~R
pr
2 dn
pr
3 4 mt
pr
5 6 mt
7 dd
In this derivation and those below, the
"dd" could occur at the end of the
previous line.
W → ~(V→~Y)
X → (V→~Y)
V→Y
(V→Y) → X
∴ ~W
1. Show ~W
2.
V→Y
3.
(V→Y) → X
4.
X
5.
X → (V→~Y)
6.
V→~Y
7.
~~(V→~Y)
8.
W → ~(V→~Y)
9.
~W
10.
pr
pr
2 3 mp
pr
4 5 mp
6 dn
pr
7 8 mt
9 dd
(W→Z) → (Z→W)
(Z→W) → ~X
P→X
~~P
∴ ~(W→Z)
1. Show ~(W→Z)
2.
~~P
3.
P
4.
P→X
5.
X
6.
~~X
7.
(Z→W) → ~X
8.
~(Z→W)
9.
(W→Z) → (Z→W)
10.
~(W→Z)
11.
Copyrighted material
pr
2 dn
pr
3 4 mp
5 dn
pr
6 7 mt
pr
8 9 mt
10 dd
Chapter One -- 49 -- Answers to the Exercises
Version of Aug 2013
Answers to the Exercises -- CHAPTER 1
SECTION 6
1. Only errors are listed.
Derivation a
All correct
Derivation b
Line 3 -- Line 1 is not available at line 3 because when line 3 is written, line 1 is still an un-cancelled show
line.
Line 7 -- No problem with the use of cd to box and cancel, but mt cannot be applied to lines 5 and 6
because line 6 does not contain the negation of the consequent of line 5; you would have to add a
line and apply double negation to line 6 first.
Derivation c
Line 2 -- You can only assume the antecedent of the conditional to be shown.
Line 3 -- While the sentence on line 3 does logically follow from line 2 and premise 1, you can't apply mp
to line 2 alone.
Line 9 -- The application of dn to line 8 is OK, but you can't end a conditional derivation on a line that does
not contain the consequent of the conditional you set out to show.
2. In each case the derivation displayed does not represent the only possible derivation; many alternate,
equally correct derivations can be given.
a.
P → (Q → (R→S))
~Q → ~R
R
∴ P→S
1. Show P → S
2.
P
3.
P → (Q → (R→S))
4.
Q → (R→S)
5.
R
6.
~~R
7.
~Q → ~R
8.
~~Q
9.
Q
10.
R→S
11.
S
12.
b.
ass cd
pr
2 3 mp
pr
5 dn
pr
6 7 mt
8 dn
4 9 mp
5 10 mp
11 cd
Q → ~(R→S)
P → (R→S)
~Q → R
∴ P→S
1. Show P → S
2.
P
3.
P → (R→S)
4.
R→S
5.
~~(R→S)
6.
Q → ~(R→S)
7.
~Q
8.
~Q → R
9.
R
10.
S
11.
Copyrighted material
ass cd
pr
2 3 mp
4 dn
pr
5 6 mt
pr
7 8 mp
4 9 mp
10 cd
Chapter One -- 50 -- Answers to the Exercises
Version of Aug 2013
Answers to the Exercises -- CHAPTER 1
c.
U → (U→V)
~R → ~(U→V)
R → ~S
∴ U → ~S
1. Show U → ~S
2.
U
3.
U → (U→V)
4.
U→V
5.
~~(U→V)
6.
~R → ~(U→V)
7.
~~R
8.
R
9.
R → ~S
10.
~S
11.
3. a
P→T
~X → ~T
~S → ~X
∴ P→S
1. Show P → S
2.
P
3.
P→T
4.
T
5.
~~T
6.
~X → ~T
7.
~~X
8.
~S → ~X
9.
~~S
10.
S
11.
b
ass cd
pr
2 3 mp
4 dn
pr
5 6 mt
pr
7 8 mt
9 dn
10 cd
T→S
Y → (S→P)
P → ~X
Y
∴ T → ~X
1. Show T → ~X
2.
T
3.
T→S
4.
S
5.
Y → (S→P)
6.
Y
7.
S→P
8.
P
9.
P → ~X
10.
~X
c
ass cd
pr
2 3 mp
4 dn
pr
5 6 mt
7 dn
pr
8 9 mp
10 cd
ass cd
pr
2 3 mp
pr
pr
5 6 mp
4 7 mp
pr
8 9 mp cd
S → ~T
~S → ~R
~R → X
Y→T
∴ ~X → ~Y
Copyrighted material
Chapter One -- 51 -- Answers to the Exercises
Version of Aug 2013
Answers to the Exercises -- CHAPTER 1
1. Show ~X → ~Y
2.
~X
3.
~R → X
4.
~~R
5.
~S → ~R
6.
~~S
7.
S
8.
S → ~T
9.
~T
10.
Y→T
11.
~Y
ass cd
pr
2 3 mt
pr
4 5 mt
6 dn
pr
7 8 mp
pr
9 10 mt cd
SECTION 7
1. Only errors are listed.
Derivation a
All correct
Derivation b
Line 3 -- The sentence on line 3 is not a premise.
Line 4 -- Line 2 is not the negation of the consequent of line 3 so mt doesn't apply. You would first have to
apply dn to line 2. Even in that case the result would be ~~S rather than ~S.
Line 5 -- "ass id" may only appear on the line immediately following a show line.
Line 7 -- The mt inference is OK, but 5 and 7 do not directly contradict so id is used incorrectly. The
derivation could be concluded on line 7 with "4 id" since 4 and 7 contradict directly.
Derivation c
Line 2 -- The sentence on the line is not the negation (or the un-negation, for that matter) of the show line.
Line 4 -- There is no way to apply mt with lines 2 and 3.
Line 6 -- 2 is not the negation of the consequent of 5 (though it should have been).
Line 7 -- There is no way that you can apply mt with lines 5 and 6; 2 and 7 don't contradict directly, so it is
premature to conclude with id.
2. In each case the derivation displayed does not represent the only possible derivation; many alternate,
equally correct derivations can be given.
a.
~Q → R
S → ~R
~S → Q
∴ Q
1. Show Q
2.
~Q
3.
~Q → R
4.
R
5.
~~R
6.
S → ~R
7.
~S
8.
~S → Q
9.
Q
10.
Copyrighted material
ass id
pr
2 3 mp
4 dn
pr
5 6 mt
pr
7 8 mp
2 9 id
Chapter One -- 52 -- Answers to the Exercises
Version of Aug 2013
Answers to the Exercises -- CHAPTER 1
b.
(P→Q) → R
S → (P→Q)
~S → R
∴ R
1.
2.
3.
4.
5.
6.
7.
8.
9.
Show R
~R
(P→Q) → R
~(P→Q)
S → (P→Q)
~S
~S → R
R
c.
~P → (R→S)
(R→S) → T
~T
Q → (R→S)
∴ ~(P → Q)
1. Show ~(P → Q)
2.
P→Q
3.
~T
4.
(R→S) → T
5.
~(R→S)
6.
Q → (R→S)
7.
~Q
8.
~P
9.
~P → (R→S)
10.
R→S
11.
ass id
pr
2 3 mt
pr
4 5 mt
pr
6 7 mp
2 8 id
ass id
pr
pr
3 4 mt
pr
5 6 mt
2 8 mt
pr
8 9 mp
5 10 id
SECTION 8
1. Only errors are listed.
Derivation a
All correct
Derivation b
Line 8 -- At this point in the derivation, line 5 is still an un-cancelled show line so line 4 can't be cited to
conclude the sub-derivation; you could, however, use the rule r (repetition) to repeat line 4 within
the sub-derivation and then use the repeated line to conclude the sub-derivation with id.
Derivation c
Line 6 -- Line 6 is not available on line 6. The problem would b resolved if we put 5 for 6.
Line 13 -- Strictly speaking there is no error here, but it was un-necessary to repeat line 2 in order to apply
id; we could have just cited "2 12 id" because line 2 is not separated from line 13 by any uncancelled show lines (only by cancelled ones).
2. In each case the derivation displayed does not represent the only possible derivation; many alternate,
equally correct derivations can be given.
a.
P → (Q→R)
Copyrighted material
Chapter One -- 53 -- Answers to the Exercises
Version of Aug 2013
Answers to the Exercises -- CHAPTER 1
S→Q
∴ S → (P→R)
1. Show S → (P→R)
2.
S
3.
Show P→R
4.
P
5.
P → (Q→R)
6.
Q→R
7.
S→Q
8.
Q
9.
R
10.
11
b.
ass cd
ass cd
pr
4 5 mp
pr
2 7 mp
6 8 mp
9 cd
3 cd
(P→Q) → Q
P→R
Q → ~Q
∴ ~(R → Q)
1. Show ~(R → Q)
2.
R→Q
3.
Show P → Q
4.
P
5.
P→R
6.
R
7.
Q
8.
(P→Q) → Q
9.
Q
10.
Q → ~Q
11.
~Q
c.
ass id
ass cd
pr
4 5 mp
2 6 mp cd
pr
3 8 mp
pr
9 10 mp 9 id
(U → V) → (W→X)
U→Z
~V → ~Z
X→Z
∴ W→Z
1. Show W → Z
2.
W
3.
Show U → V
4.
U
5.
U→Z
6.
Z
7.
~~Z
8.
~V → ~Z
9.
~~V
10.
V
11.
(U → V) → (W→X)
12.
W→X
13.
X
14.
X→Z
15.
Z
Copyrighted material
ass cd
ass cd
pr
4 5 mp
6 dn
pr
7 8 mt
9 dn cd
pr
3 11 mp
2 12 mp
pr
13 14 mp cd
Chapter One -- 54 -- Answers to the Exercises
Version of Aug 2013
Answers to the Exercises -- CHAPTER 1
SECTION 9
1. a
P→R
Q → ~R
~Q → Q
∴ P→Q
1.
2.
3.
4.
5.
6.
7.
8.
9.
Show P → Q
P
P→R
R
~~R
Q → ~R
~Q
~Q → Q
Q
ass cd
pr
2 3 mp
4 dn
pr
5 6 mt
pr
7 8 mp cd
The only change was to conclude with cd instead of id.
b
1.
2.
3.
4.
5.
6.
7.
8.
9.
Q→U
Q → ~U
R→Q
R
∴ P
Show P
~P
R
R→Q
Q
Q→U
U
Q → ~U
~U
ass id
pr
pr
3 4 mp
pr
5 6 mp
pr
5 8 mp 7 id
The only change was to add an assumption for id.
c
U → (V→W)
X→U
~X → W
∴ V→W
1. Show V → W
2.
~(V→W)
3.
U → (V→W)
4.
~U
5.
X→U
6.
~X
7.
~X → W
8.
W
9.
Show V → W
10.
V
11.
W
12.
Only added lines 9-12.
Copyrighted material
ass id
pr
2 3 mt
pr
4 5 mt
pr
6 7 mp
ass cd
8 r cd
2 9 id
Chapter One -- 55 -- Answers to the Exercises
Version of Aug 2013
Answers to the Exercises -- CHAPTER 1
2. a
P→R
Q → ~R
~Q → Q
∴ P→Q
1.
2.
3.
4.
5.
6.
Show P → Q
P
R
~~R
~Q
Q
b
1.
2.
3.
4.
5.
1.
2.
3.
4.
5.
Q→U
Q → ~U
R→Q
R
∴ P
Show P
~P
Q
U
~U
c
ass cd
2 pr1 mp
3 dn
4 pr2 mt
5 pr3 mp cd
ass id
pr4 pr3 mp
3 pr1 mp
3 pr2 mp 4 id
U → (V→W)
X→U
~X → W
∴ V→W
Show V → W
~(V→W)
~U
~X
W
ass id
2 pr1 mt
3 pr2 mt
4 pr3 mp cd
3
P
Q
∴ P→Q
1.
2.
Show P → Q
Q
pr2 cd
P
~Q
∴ ~(P→Q)
1.
2.
3.
4.
Show ~(P→Q)
P→Q
Q
~Q
Copyrighted material
ass id
pr1 2 mp
pr2 3 id
Chapter One -- 56 -- Answers to the Exercises
Version of Aug 2013
Answers to the Exercises -- CHAPTER 1
~P
Q
∴ P→Q
1.
2.
Show P → Q
Q
pr2 cd
~P
~Q
∴ P→Q
1.
2.
3.
Show P → Q
P
~P
ass cd
pr1 2 id
SECTION 10
1.
S
(R→S) → W
∴ W
1.
2.
3.
4.
Show W
Show R → S
S
W
2.
P → (S→R)
P → (W→S)
W→P
∴ W→R
1.
2.
3.
4.
5.
6.
7.
Show W → R
W
P
S→R
W→S
S
R
3.
(P→Q) → S
S→T
~T → Q
∴ T
1.
2.
3.
4.
5.
6.
7.
Show T
~T
Q
Show P→Q
Q
S
T
SECTION 11
Copyrighted material
pr1 cd
2 pr2 mp dd
as cd
2 pr3 mp
3 pr1 mp
3 pr2 mp
2 5 mp
4 6 mp cd
ass id
2 pr3 mp
3 r cd
4 pr1 mp
6 pr2 mp 2 id
Derivations for theorems are not given here.
Chapter One -- 57 -- Answers to the Exercises
Version of Aug 2013
Answers to the Exercises -- CHAPTER 1
SECTION 12
1. a
X → ~(Y → Z)
∴ (Y → Z) → ~X
1.
2.
3.
Show (Y → Z) → ~X
(X → ~(Y → Z)) → ((Y → Z) → ~X)
(Y → Z) → ~X
b
c
1.
2.
3.
4.
5.
6.
7.
8.
9.
T6
2 pr1 mp
3 pr2 mp dd
~(R → (S→T))
R→P
P → (Q → (S→T))
∴ ~Q
Show ~Q
Q
~(R → (S→T)) → R
R
P
Q → (S→T)
S→T
~(R → (S→T)) → ~(S→T)
~(S→T)
d
1.
2.
3.
4.
R → (~P→S)
R → ~P
∴ R→S
Show R → S
(R → (~P→S)) → ((R→~P) → (R→S))
(R→~P) → (R→S)
R→S
1.
2.
3.
4.
T14
2 pr1 mp dd
ass id
T21
3 pr1 mp
4 pr2 mp
5 pr3 mp
2 6 mp
T22
8 pr1 mp 7 id
Q→R
R→S
∴ Q→S
Show Q → S
(Q→R) → ((R→S) → (Q→S))
(R→S) → (Q→S)
Q→S
Copyrighted material
T4
2 pr1 mp
3 pr2 mp dd
Chapter One -- 58 -- Answers to the Exercises
Version of Aug 2013
9/17/2013
CHAPTER 2 SECTION 1
Chapter Two
Sentential Logic with 'and', 'or', if-and-only-if'
1 SYMBOLIC NOTATION
In this chapter we expand our formal notation by adding three two-place connectives, corresponding
roughly to the English words 'and', 'or' and 'if and only if':
∧
∨
↔
and
or
if and only if
Conjunction: The first of these, '∧', is the conjunction sign; it has the same logical import as 'and'. It
goes between two sentences to form a complex sentence which is true if both of the parts (called
'conjuncts') are true, and is otherwise false:
□
○
(□∧○)
T
T
T
T
F
F
F
T
F
F
F
F
Disjunction: The disjunction sign, '∨', makes a sentence that is true in every case except when its parts
(its disjuncts) are both false. This corresponds to one use (the "inclusive" use) of 'or' in English:
□
T
T
F
F
○
T
F
T
F
(□ ∨ ○)
T
T
T
F
Biconditional: The biconditional sign, '↔', states that both of the parts making it up (its constituents)
are the same in truth value. It works like this:
□
T
T
F
F
○
T
F
T
F
(□ ↔ ○)
T
F
F
T
Each of these new connectives behaves syntactically just like the conditional sign, '→': you make a bigger
sentence out of two sentences plus a pair of parentheses:
(□∧○)
(□ ∨ ○)
(□ ↔ ○)
Our expanded definition of a sentence in official notation is now:
Chapter Two SYMBOLIC SENTENCES
•
•
•
Any capital letter between 'P' and 'Z' is a symbolic sentence.
(Numerical subscripts may also be used, as 'P3', 'Q24'.)
If □ is a symbolic sentence, so is ~□
If □ and ○ are symbolic sentences, so are (□→○), (□∧○), (□ ∨ ○), and (□ ↔ ○).
Nothing is a symbolic sentence for purposes of chapter 2 unless it can be generated by
the clauses given above.
Copyrighted material
Chapter 2 -- 1
Version of Aug 2013
9/17/2013
CHAPTER 2 SECTION 1
As before, we allow ourselves informally to omit the outer parentheses when the sentence occurs alone
on a line. It is also customary (and convenient) to omit parentheses around conjunctions or disjunctions
when they are combined with a conditional or biconditional sign. The sentence:
P∧Q → R
is to be considered to be an informally worded conditional whose antecedent is a conjunction:
(P∧Q) → R
If we want to make a conjunction whose second conjunct is a conditional, we must use parentheses
around the parts of the conditional:
P ∧ (Q→R)
Likewise, this sentence:
P ↔ Q∨R
is an informally written biconditional whose second constituent is a disjunction:
P ↔ (Q∨R)
If we wish to write a disjunction whose first disjunct is a biconditional, we need to use parentheses around
the biconditional:
(P↔Q) ∨ R.
Finally, we may use two or more conjunction signs or disjunction signs (but not a mix of conjunctions with
disjunctions) as abbreviations for what you get by restoring the parentheses by grouping the left parts
together, so that: 'P ∧ Q ∧ R' is an abbreviation for '(P∧Q) ∧ R'.
Informal Conventions
Outermost parentheses may be omitted.
Conjunction signs or disjunction signs may be used with conditional signs or
biconditional signs with the understanding that this is short for a conditional or
biconditional which has a conjunction or disjunction as a part. For example:
P∨Q → R
is informal notation for
(P∨Q) → R
P ↔ Q∧R
is informal notation for
P ↔ (Q∧R)
Repeated conjuncts or disjuncts without parentheses are short for the result of putting
parentheses around the part to the left of the last conjunction or disjunction sign. For
example:
P∨Q∨R
is informal notation for
(P∨Q) ∨ R
P∧Q∧R
is informal notation for
(P∧Q) ∧ R
Sentences with the new connectives may be parsed as we did in the previous chapter:
P∧Q → R
2
P∧Q
R
2
P
Q
Copyrighted material
P ↔ Q∨R
2
P
Q∨R
2
Q
R
Chapter 2 -- 2
Version of Aug 2013
9/17/2013
CHAPTER 2 SECTION 1
~(P∧Q) → (R ↔ P∨Q)
2
~(P∧Q)
R ↔ P∨R
|
2
P∧Q
R
P∨R
2
2
P
Q
P
R
~~(R ↔ (P → ~Q))
|
~(R ↔ (P → ~Q))
|
R ↔ (P → ~Q)
2
R
P→~Q
2
P
~Q
|
Q
Determining Truth Values Using such parsings, there is a mechanical way to determine whether any
given sentence is true or false if you know the truth values of the sentence letters making it up. First,
make a parse tree as above by taking the sentences on any given line and writing their immediate parts
below them. A parse tree for '(P∧Q) → (P∨R)' is:
(P ∧ Q) → (P ∨ R)
2
(P ∧ Q)
(P ∨ R)
2
2
P
Q
P
R
Then write the truth values of the sentence letters below them. For example, if P and Q are both true but
R false, you would have:
(P ∧ Q) → (P ∨ R)
2
(P ∧ Q)
(P ∨ R)
2
2
P
Q
P
R
T
T
T
F
Then go up the parse tree, placing a truth value under the major connective of each sentence based on
the truth values of its parts given below. For example, the truth value under '(P ∧ Q)' would be 'T' because
it is a conjunction, and both of its parts are T:

(P ∧ Q) → (P ∨ R)
2
(P ∧ Q)
(P ∨ R)
T
2
2
P
Q
P
R
T
T
T
F
Filling in the remaining parts gives you a truth value for the whole sentence at the top:
Copyrighted material
Chapter 2 -- 3
Version of Aug 2013
9/17/2013
CHAPTER 2 SECTION 1
( (P ∧ Q) → (P ∨ R) )
T
2
(P ∧ Q)
(P ∨ R)
T
T
2
2
P
Q
P
R
T
T
T
F
Sometimes not all of the parse tree needs to be filled out; this happens when partial information below a
sentence is sufficient to decide its truth value. In the example just given it is not necessary to figure out
the truth value of '(P ∧ Q)', since the conditional on the top line is determined to be true based on the
information that '(P ∨ R)' is true. So the following parse tree is sufficient to show that the main sentence is
true if the sentence letters have the indicated truth values:
(P ∧ Q) → (P ∨ R)
T
2
(P ∧ Q)
(P ∨ R)
T
2
2
P
Q
P
R
T
F
EXERCISES
1. For each of the following state whether it is a sentence in official notation, or a sentence in informal
notation, or not a sentence at all. If it is a sentence, parse it as indicated above.
a.
b.
c.
d.
e.
f.
g.
h.
i.
P↔Q→R
~Q↔~R
~(Q↔R)
P∧Q∨R
(P→Q) ∨ (R→~Q)
P ↔ (Q∧R) → Q
P∧Q → (Q→R∨Q)
P ↔ (P↔Q∧R)
P ∨ (Q→P)
2. If 'P' and 'Q' are both true and 'R' is false, what are the truth values of the official or informal sentences
in 1? (Use the parses that you give in 1 to guide the determination of truth values.)
Copyrighted material
Chapter 2 -- 4
Version of Aug 2013
CHAPTER 2 SECTION 2
2 ENGLISH EQUIVALENTS OF THE CONNECTIVES
Conjunctions: The word 'and' is equivalent to the symbol '∧'. There are other locutions of English that
may also be equivalent to '∧', although they are sometimes used to communicate something additional.
For example:
The book is short, and it is interesting
The book is short, but it is interesting
The book is short, although it is interesting
The book is short, even though it is interesting
The book is short; it is interesting
Some of these sentences suggest that if a book is short, you probably won't find it interesting. But all that
they literally say is that it is both short and interesting. If it isn't short, what you have said is false, and if it
isn't interesting then what you have said is false, but if it is both short and interesting, what you have said
is true, even if possibly misleading.
Conjunctions: □ ∧ ○
□ and ○
both □ and ○
□ but ○
□ although ○
although □, ○
□ even though ○
even though □, ○
□;○
In certain cases, use of a relative pronoun is logically equivalent to a use of '∧': the sentence 'Maria, who
was late, greeted the vice-counsel' is equivalent to 'Maria was late ∧ Maria greeted the vice-counsel'.
Disjunctions: The English word 'or' can be taken in two ways: inclusively or exclusively. If you are asked
to contribute food or money, you will probably take this as saying that you may contribute either or both;
the invitation is inclusive. But if a menu says that you may have soup or salad the normal interpretation is
that you may have either, but not both; the offer is exclusive. The difference in logical import appears in
the first row here:
□
T
T
F
F
○
T
F
T
F
(□ inclusive-or ○)
T
T
T
F
(□ exclusive-or ○)
F
T
T
F
If the English 'or' can be read either inclusively or exclusively, we will need to have a convention for how to
interpret it when it is used in exercises. Our convention will be that 'or' is always meant inclusively when it
is used in problems and examples in this text. That is, it coincides in logical import with our disjunction
sign '∨'.
A common synonym of 'or' is 'unless'. The sentence 'Wilma will leave unless there is food' is false if there
is no food but Wilma doesn’t leave; otherwise it is true, just like 'or' when read inclusively.
Disjunctions: □ ∨ ○
□ or ○
either □ or ○
□ unless ○
Biconditionals: We will see below that a biconditional sign is equivalent to two conditionals made from
its constituents. The sentence '(□ ↔ ○)' is equivalent to:
Copyrighted material
Chapter 2 -- 5
Version of Aug 2013
CHAPTER 2 SECTION 2
(□ → ○) ∧ (○ → □)
This can be read in English as '○ if □, and ○ only if □'; thus it is often pronounced 'if and only if'. The
English phrase 'just in case' or 'exactly in case' are sometimes used to state the equivalence of two
claims; the biconditional can be used to symbolize them:
The game will be called off just in case it rains:
The game will be played exactly in case it is sunny:
Q↔R
P↔S
Biconditionals: □ ↔ ○
□ if and only if ○
□ exactly on condition that ○
□ just in case ○
EXERCISES
1. For each of the following sentences say which symbolic sentence it is equivalent to.
a. It will rain, but the game will be played anyway.
R∧P
R→P
R↔P
b. Willa drove or got a ride
W∨R
W↔R
c. Robert, who didn't get a ride, was tardy
~R → T
~R ∧ T
d. It rained; the sell-a-thon was called off
R↔S
R∧S
e. The quilting bee will be called off just in case it rains
Q∧R
Q↔R
Q→R
R→Q
2. Symbolize each of the following using this translation scheme:
S
V
R
Q
T
Sally will walk
Veronica will give Sally a ride
It will rain
Barbara will come with Quincy
Barbara will come with Tom
a. Sally will walk or Veronica will give her a ride.
b. Exactly on condition that it rains will Sally walk
c. Although it will rain, Sally will walk
d. Barbara will come with Quincy or Tom
e. Barbara will come with Quincy; Sally will walk
3. What are the truth values of the sentences in 2 when all of the simple sentences are false?
Copyrighted material
Chapter 2 -- 6
Version of Aug 2013
CHAPTER 2 SECTION 3
3 COMPLEX SENTENCES
Complex sentences of English generally translate into complex sentences of the logical notation. As
usual, it is important to be clear about the grouping of clauses in the English sentence.
The following sentence is a simple conjunction:
P∧Q
Polk and Quincy were presidents
The following sentence is also a conjunction, one of whose conjuncts is a negation:
Polk, but not Quincy, was a president.
P ∧ ~Q
This is a negation of a conjunction:
Not both Polk and Quincy were presidents.
~(P ∧ Q)
This is a simple disjunction:
Either Polk or Quincy was president.
P∨Q
This is a complex sentence, with at least two different but equivalent symbolizations.
Neither Polk nor Quincy was president.
One symbolization is the negation of 'Either Polk or Quincy was president; in this symbolization 'neither'
means 'not either': ~(P ∨ Q). An equivalent symbolization is a conjunction of negations; 'neither P nor Q'
is equivalent to "not P and not Q": ~P ∧ ~Q
The fundamental principles for our new connectives are:
and, or, if and only if
When any of these expressions occurs between sentences, it gives rise to a
conjunction, disjunction, or biconditional. The constituents of the conjunction,
disjunction, or biconditional are symbolizations of sentences immediately to the
left and to the right of 'and', 'or', or 'if and only if'.
When 'either' occurs with 'or', the symbolization of the expression enclosed
between 'either' and 'or' is a disjunct. Likewise, When 'both' occurs with 'and',
the symbolization of the expression enclosed between 'both' and 'and' is a
conjunct.
'neither □ nor ○' is equivalent to 'not (either □ or ○)'.
As in chapter 1, these principles do not eliminate all ambiguity. The sentence 'Wilma will leave and Steve
will stay or Tom will dance' is ambiguous between these two symbolizations:
W & (S∨T)
(W&S) ∨ T
The use of 'either' will sometimes disambiguate; the only symbolization of 'Wilma will leave and either
Steve will stay or Tom will dance' is:
W & (S∨T)
This is because 'either' and 'or' exactly enclose 'Steve will stay', and so 'S' must be a disjunct. But it is not
a disjunct in '(W&S) ∨ T'.
Commas play their usual role of grouping items on each side. The sentence 'Wilma will leave and Steve
will stay, or Tom will dance' has only the symbolization:
Copyrighted material
Chapter 2 -- 7
Version of Aug 2013
CHAPTER 2 SECTION 3
(W&S) ∨ T
Conjunction and disjunction signs inside of sentences: Sometimes 'and' and 'or' occur within
sentences, as in:
Wilma sang and danced
Tom or Sam left
In such cases you need to fill in a missing part to get a sentence that we already know how to symbolize.
Sometimes 'and' or 'or' occurs inside a simple sentence, where only the subject is
conjoined or disjoined, and there is a single predicate, or only the predicate is conjoined
or disjoined, and there is a single subject. If you fill in a copy of the shared part, you will
get a synonymous sentence that we already know how to symbolize.
These are some examples:
Wilma sang and danced
Tom or Sam left
Wilma sang and [Wilma] danced
Tom [left] or Sam left
If there is a 'both' or an 'either, it ends up on the front:
Both Tom and Sam left
Either Tom or Sam left
Both Tom [left] and Sam left
Either Tom [left] or Sam left
Wilma both sang and danced
Wilma either sang or danced
Both Wilma sang and [Wilma] danced
Either Wilma sang or [Wilma] danced.
There may also be a 'not' after the compound subject, or before a compound predicate. If the negation is
after a compound subject, it forms part of the predicate, and it is filled in with that predicate:
Wilma or Veronica didn't sing
Wilma [didn't sing] or Veronica didn't sing.
If the negation is before a compound predicate, it yields a negation sign that applies to the whole
compound:
Wilma didn't sing or dance
Wilma didn't sing and dance
~ (Wilma sang or danced)
~ (Wilma sang and danced)
The parts inside the parentheses are then expanded as usual:
Wilma didn't sing or dance
Wilma didn't sing and dance
~ (Wilma sang or [Wilma] danced)
~ (Wilma sang and [Wilma] danced)
Compounds within simple sentences affect how sentences are grouped after symbolization:
When connectives occur inside otherwise simple sentences, the symbolizations of the
sentences form a unit.
For example, the sentence 'Ruth tap-dances or sings and she plays the clarinet' must be grouped like this:
(T ∨ S) & P
This is because the disjunction with 'T' and 'S' must be a unit. In 'Ruth tap-dances or she sings and plays
the clarinet' the opposite happens; you must have:
T ∨ (S & P)
because the conjunction with 'S' and 'P' must form a unit.
Copyrighted material
Chapter 2 -- 8
Version of Aug 2013
CHAPTER 2 SECTION 3
Synonyms of 'and', 'or', and 'if and only it' are subject to the conditions described above.
Here are some illustrations:
If neither Wilma nor Sally attends, either Robert or Peter will be bored.
If neither Wilma [attends] nor Sally attends, either Robert [will be bored] or Peter will be bored.
If neither W nor S, either R or P
~(W∨S) → (R∨P)
The 'neither' and the 'either' made units, and the comma was redundant.
A slightly more complex case:
If neither Wilma nor Sally attends, either Robert or Peter, but not Tom, will be bored.
If neither Wilma [attends] nor Sally attends, either Robert [will be bored] or Peter [will be bored],
but Tom will not be bored.
If neither W nor S, either R or P, but not T
~(W∨S) → (R∨P) & ~T
Here the 'either' made a unit, so the 'but ' could not take 'Peter would be bored' as its left conjunct. That
is, the symbolization could not be this:
~(W∨S) → (R ∨ (P&~T))
Likewise, the original phrase 'either Robert or Peter, but not Tom, will be bored' consists of three simple
sentences all sharing the 'will be bored', so it could not be pulled apart, as in:
(~(W∨S) → (R∨P)) & ~T
Some additional examples:
Either Robert or Tom will attend, but not both
Either Robert [will attend] or Tom will attend, but not both [will attend]
Either R or T, but not R and T
(R∨T) ∧~(R∧T)
Robert will attend if Sally does, but she won't attend if neither Tom nor Wilma attend.
Robert will attend if Sally does [attend], but she won't attend if neither Tom [attends] nor Wilma
attends.
R if S, but not S if neither T nor W
(S→R) ∧(~(T∨W) → ~S)
Neither Sally nor Robert will run, but if either Tom or Quincy run, Veronica will win.
Neither S nor R, but if either T or Q, V
~(S∨R) ∧(T∨Q → V).
Given that Sally and Robert won't both run, Tom will run exactly if Q does.
Given that not both S and R, T exactly if Q.
~(S∧R) → (T↔Q)
A variety of English expressions that we have not mentioned affect how a sentence is to be symbolized.
Examples:
Quincy will whistle if Reggie sings without Susan singing or Susan sings without Reggie, but he
won't whistle if they both sing
Copyrighted material
Chapter 2 -- 9
Version of Aug 2013
CHAPTER 2 SECTION 3
Q if R and not S or S and not R, but not Q if S and R
((R∧~S) ∨ (S∧~R) → Q) ∧(S∧R → ~Q)
Here 'Reggie sings without Susan singing' means that Reggie sings and Susan doesn't sing.
If Sally runs, Rob will run, in which case Theodore will leave
(S→R) ∧ (R→T)
Here ' in which case' means "if Rob runs".
If a symbolization of a sentence is a correct one, then it and the English sentence being symbolized must
agree in truth value no matter what truth values the simple sentences have. If they agree for every
assignment of truth values, then the symbolization is correct. If not, it is incorrect. (To tell whether an
English sentence is true or false given a specification of truth values for its simple parts you must rely on
your understanding of English. To tell whether a symbolic sentence is true or false given the truth values
of its sentential letters, you parse it and figure out its truth value as in section 1.)
EXERCISES
1. If 'P' is true and both 'Q' and 'R' are false, what are the truth values of the following? (In answering, give
a parse tree for the sentence.)
a. ~(P ∨ (Q ∧ R))
b. ~P ∨ (Q ∧ R)
c. ~(P ∨ R) ↔ ~P ∨ R
d. ~Q ∧ (P ∨ (Q↔R))
e. P → (~Q ↔ (~R → Q))
For questions 2 and 3, use this translation scheme:
V
Veronica will leave
W
William will leave
Y
Yolanda will leave
2. For each of the following say which of the proposed translations is correct.
a. Veronica won't leave if and only if William won’t leave
~(V ↔ ~W)
~V ↔ ~W
V ↔ ~~W
b. William and Veronica will both leave if Yolanda does, provided that Veronica doesn’t
Y∧~V → W∧V
(Y→W∧V) → ~V
~V → (Y→W∧V)
c. Unless Yolanda leaves, Veronica or William will leave
Y ∨ (W ∨ V)
Y → (W ∨ V)
Y↔W∧V
d. Either Yolanda leaves and Veronica doesn't, or Veronica leaves and William doesn’t
(Y ↔ ~V) ∨ (V ↔ ~W)
(Y ∧ ~V) ∨ (V ∧ ~W)
Y ∧ ~V ↔ V ∧ ~W
Copyrighted material
Chapter 2 -- 10
Version of Aug 2013
CHAPTER 2 SECTION 3
3. For each of the following produce a correct symbolization
a. Only if Veronica doesn't leave will William leave, or Veronica and William and Yolanda will all
leave
b. If neither William nor Veronica leaves, Yolanda won't either
c. If William will leave if Veronica leaves, then he will surely leave if Yolanda leaves
d. Neither William nor Veronica nor Yolanda will leave
4. What are the truth values of 3a-d if Veronica leaves but neither William nor Yolanda leaves?
For question 5 use this translation scheme:
R
W
Q
Sally will run
Sally will win
Sally will quit
5. For each of the following produce a correct symbolization
a. Sally will run and win unless she quits
b. Sally will win exactly in case she runs without quitting
c. Sally, who will run, will win if she doesn't quit
d. Sally will run and quit, but she will win anyway
Copyrighted material
Chapter 2 -- 11
Version of Aug 2013
CHAPTER 2 SECTION 4
4 RULES
Each new connective comes with two new rules. As earlier, it should be obvious from the truth-table
descriptions of each connective that instances of these rules are formally valid arguments.
Conjunction rules:
Rule s (simplification)
□∧○
∴ □
or
Rule adj (adjunction)
□∧○
∴○
□
○
∴ □∧○
Disjunction rules:
Rule add (addition)
□
∴ □∨○
or
Rule mtp (modus tollendo ponens)
□
∴○∨□
□∨○
~○
∴ □
Biconditional rules:
Rule bc (biconditional-to-conditional)
□↔○
∴ □→○
or
or
□∨○
~□
∴○
Rule cb (conditionals-to-biconditional)
□↔○
∴○→□
□→○
○→□
∴ □↔○
Simplification indicates that if you have a conjunction, you may infer either conjunct. For example, both
of these valid arguments are instances of rule s:
P∧W
∴ P
by rule s
∴ W
by rule s
Polk was a president and so was Whitney
∴ Polk was a president
∴ Whitney was a president
Adjunction indicates that if you have any two sentences, you may infer their conjunction, in either order.
For example, these valid arguments are instances of rule adj:
Polk was a president
Whitney was a president
∴ Polk was a president and so was Whitney
∴Whitney was a president and so was Polk
P
W
∴ P∧W
∴ W∧P
by rule adj
by rule adj
This derivation illustrates how the conjunction rules are used:
P∧Q
∴ Q∧P
1. Show Q ∧ P
2.
3.
4.
Q
P
Q∧P
pr1 s
pr1 s
2 3 adj dd
Addition indicates that from any sentence you may infer its disjunction with any other sentence.
Polk was a president
∴ Polk was a president or Whitney was
∴ Whitney was a president or Polk was
Copyrighted material
P
∴ P∨W
∴ W∨P
Chapter 2 -- 12
by rule add
by rule add
Version of Aug 2013
CHAPTER 2 SECTION 4
Rule add lets you add any disjunct, no matter how irrelevant. So from 'Cynthia left' you may infer 'Cynthia
left ∨ Fido barked'. This is legitimate because '∨' is used inclusively, and all that you need for a disjunction
to be true is that either disjunct be true. So if 'Cynthia left' is true, 'Cynthia left ∨ Fido barked' must be true
too.
Modus tollendo ponens indicates that from a disjunction and the negation of one of its disjuncts you may
infer the other disjunct.
Polk was a president or Whitney was
Whitney wasn't a president
∴ Polk was a president
P∨W
~W
∴ P
by rule mtp
Polk was a president or Whitney was
Polk wasn't a president
∴ Whitney was a president
P∨W
~P
∴ W
by rule mtp
Note that the following is not an instance of modus tollendo ponens:
Whitney was a president or Truman was
Truman was a president
∴ Whitney wasn't a president
W∨T
T
∴ ~W
For mtp you need the negation of a disjunct. In the case given, if 'T' and 'W' were both true, then the
argument would have true premises and a false conclusion.
Here is a derivation illustrating the disjunction rules. It is a derivation for this argument:
P
R ∨ ~P
∴ R∨S
1. Show R ∨ S
2.
3.
4.
~~P
R
R∨S
pr1 dn
2 pr2 mtp
3 add dd
Biconditional-to-conditional indicates that from a biconditional you may infer either of the corresponding
conditionals:
Polk was a president if and only if Whitney was
∴ If Polk was a president, so was Whitney
∴ If Whitney was a president, so was Polk
P↔W
∴ P→W
∴ W→P
by rule bc
by rule bc
Conditionals-to-biconditional indicates that from two conditionals where the antecedent of one is the
consequent of the other, and vice versa, you may infer a bicondtional containing the parts of the
conditionals:
If Polk was a president, so was Whitney
If Whitney was a president, so was Polk
∴ Polk was a president if and only if Whitney was
P→W
W→P
∴ P↔W
by rule cb
Here are two more derivations using our new rules:
S∧P
P∨Q → ~R
Q∨R
∴ Q
Copyrighted material
Chapter 2 -- 13
Version of Aug 2013
CHAPTER 2 SECTION 4
1. Show Q
2.
3.
4.
5.
P
P∨Q
~R
Q
pr1 s
2 add
3 pr2 mp
4 pr3 mtp dd
R ↔ ~P
~Q ↔ R
∴ P↔Q
1. Show P ↔ Q
2.
Show P → Q
3.
4.
5.
6.
7.
8.
9.
P
~~P
R → ~P
~R
~Q → R
~~Q
Q
10.
Show Q → P
11.
12.
13.
14.
15.
16.
17.
Q
~~Q
R → ~Q
~R
~P → R
~~P
P
18.
P↔Q
ass cd
3 dn
pr1 bc
4 5 mt
pr2 bc
6 7 mt
8 dn cd
ass cd
11 dn
pr2 bc
12 13 mt
pr1 bc
14 15 mt
16 dn cd
2 10 cb dd
EXERCISES
1. For each of the following arguments, say which rule it is an instance of (or say "none").
a.
P ∨ ~Q
Q
∴ P
b.
~P ∧ Q
∴ ~P
c.
~~(P→Q)
∴ P→Q
d.
~P∨Q
~Q
∴ ~P
e.
~P → ~Q
~Q → ~P
∴ ~Q ↔ ~P
f.
P∨Q
~R
∴ P
g.
~~P ↔ R
∴ R → ~~P
h.
Q
∴ ~P ∨ Q
i.
P∨Q
∴ Q
2. Given the sentences below, say what can be inferred in one step by s, mtp, bc, cb using all of the
premises.
a.
~W → ~X
~X → ~W
∴ ?
b.
~W ∨ ~X
~~X
∴ ?
c.
W→X
~W
∴ ?
d.
~W ∧ ~X
∴ ?
e.
W ↔ ~X
∴ ?
f.
W∨X
∴ ?
Copyrighted material
Chapter 2 -- 14
Version of Aug 2013
CHAPTER 2 SECTION 5
5 SOME DERIVATIONS USING RULES S, ADJ, CB
Since there are new connectives it is useful to expand our strategy hints from Chapter 1:
Additional Strategy Hints
When trying to derive a conjunction, derive the conjuncts and then use adj.
When trying to derive a biconditional, derive the corresponding conditionals and use cb.
These strategy hints will be put to use below, as we extend our list of Theorems from Chapter 1.
Theorem 24 is the commutative law for conjunction; it says that turning the conjuncts of a sentence
around produces a logically equivalent sentence:
T24
P∧Q ↔ Q∧P
"commutative law for conjunction"
This is easy to derive if you follow the last two strategy hints. You will be deriving a biconditional, so you
will try to derive both conditionals: P∧Q → Q∧P and Q∧P → P∧Q and then combine them using rule cb.
While deriving each conditional you will derive a conjunction by deriving its conjuncts and then using rule
adj. Rule s is used whenever you want to get one of the conjuncts of an existing conjunction alone.
1. Show P∧Q ↔ Q∧P
2.
3.
4.
5.
6.
7.
Show P∧Q → Q∧P
P∧Q
P
Q
Q∧P
ass cd
3s
3s
4 5 adj cd
Show Q∧P → P∧Q
8.
9.
10.
11.
Q∧P
Q
P
P∧Q
ass cd
8s
8s
9 10 adj cd
12.
P∧Q ↔ Q∧P
2 7 cb dd
The next theorem is the associative law for conjunction; it says that regrouping successive conjuncts
produces a logically equivalent sentence. The strategy here is the same as that above: to derive the
biconditional you derive the corresponding conditionals, and use rule cb. In deriving the conditionals you
derive conjunctions using rule adj. Again, rule s is used to simplify conjunctions that you already have.
Copyrighted material
Chapter 2 -- 15
Version of Aug 2013
CHAPTER 2 SECTION 5
T25 P∧ (Q∧R) ↔ (P∧Q)∧ R
"associative law for conjunction"
1. Show P∧ (Q∧R) ↔ (P∧Q)∧ R
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
Show P∧ (Q∧R) → (P∧Q)∧ R
P∧ (Q∧R)
P
Q∧R
Q
R
P∧Q
(P∧Q) ∧R
ass cd
3s
3s
5s
5s
4 6 adj
7 8 adj cd
Show (P∧Q)∧ R → P∧ (Q∧R)
(P∧Q)∧ R
R
P∧Q
P
Q
Q∧R
P∧ (Q∧R)
P∧ (Q∧R) ↔ (P∧Q)∧ R
ass cd
3s
3s
5s
5s
4 6 adj
7 8 adj cd
2 10 cb dd
The next theorem is T26:
T26
(P→Q) ∧(Q→R) → (P→ R)
"hypothetical syllogism"
Notice that T26 and T4 from the previous chapter are both called "hypothetical syllogism".
T4 (P→Q) → ((Q→R) → (P→R))
"hypothetical syllogism"
These two theorems are closely related. They are related to one another as the following two patterns:
□∧○→△
□ → (○ → △)
where each theorem has 'P→Q' in place of □, 'Q→R' in place of ○, and 'P→R' in place of △. Our next
theorem says that these two patterns are equivalent.
T27
(P∧Q → R) ↔ (P → (Q→R))
"exportation"
The derivation of this theorem is also relatively straightforward: derive two conditionals and put them
together by rule cb. Each conditional itself has conditionals as parts, so the derivation calls for two
conditional subderivations (one of which itself contains another conditional subderivation).
Copyrighted material
Chapter 2 -- 16
Version of Aug 2013
CHAPTER 2 SECTION 5
1. Show (P∧Q → R) ↔ (P→ (Q→R))
2.
Show (P∧Q → R) → (P→ (Q→R))
3.
4.
P∧Q → R
Show P→ (Q→R)
ass cd
5.
6.
P
Show Q → R
ass cd
7.
8.
9.
Q
P∧Q
R
ass cd
5 7 adj
3 8 mp cd
10.
6 cd
11.
4 cd
12. Show (P→ (Q→R) → (P∧Q → R)
13.
14.
P → (Q→R)
Show P∧Q → R
ass cd
P∧Q
P
Q→R
Q
R
15.
16.
17.
18.
19.
ass cd
15 s
13 16 mp
15 s
17 18 mp cd
20.
14 cd
21. (P∧Q → R) ↔ (P→ (Q→R))
2 12 cb dd
This derivation is complex. It may be useful to see how we might think up how to construct it. First, our
main strategy is to derive a biconditional by deriving two conditionals. So our plan predicts that the
derivation will have this overall structure:
1. Show (P∧Q → R) ↔ (P→ (Q→R))
2.
(P∧Q → R) → (P→ (Q→R))
12. (P→ (Q→R) → (P∧Q → R)
21. (P∧Q → R) ↔ (P→ (Q→R))
2 12 cb dd
Line 2 will require a conditional derivation, and so will line 12. So the completed derivation will take this
form:
1. Show (P∧Q → R) ↔ (P→ (Q→R))
2.
3.
Show (P∧Q → R) → (P→ (Q→R))
P∧Q → R
ass cd
P → (Q→R)
xxx cd
12. Show (P→ (Q→R) → (P∧Q → R)
13.
P → (Q→R)
ass cd
P∧Q → R
xxx cd
21. (P∧Q → R) ↔ (P→ (Q→R))
Copyrighted material
2 12 cb dd
Chapter 2 -- 17
Version of Aug 2013
CHAPTER 2 SECTION 5
Lines 3-11 and 14-20 are taken up with completing the subderivations. Each of these itself uses a
conditional subderivation, giving the following structure:
1. Show (P∧Q → R) ↔ (P→ (Q→R))
2.
Show (P∧Q → R) → (P→ (Q→R))
3.
4.
P∧Q → R
Show P→ (Q→R)
5.
ass cd
P
ass cd
10.
xxx cd
11.
4 cd
12. Show (P→ (Q→R) → (P∧Q → R)
13.
14.
P → (Q→R)
Show P∧Q → R
15.
ass cd
P∧Q
ass cd
19.
xxx cd
20.
14 cd
21. (P∧Q → R) ↔ (P→ (Q→R))
2 12 cb dd
The rest of the work is filling in the remaining subderivations. It is often useful to develop a derivation as
we did here by first sketching its overall structure, and then flesh it out with details afterwards.
EXERCISES
1. Produce derivations for theorems T28-T30, T33, T36-37, which are included among the theorems
stated here:
T28
(P∧Q → R) ↔ (P∧~R → ~Q)
T29
(P→ Q∧R) ↔ (P→Q) ∧(P→R)
T30
(P→Q) → (R∧P → R∧Q)
T31
(P→Q) → (P∧R → Q∧R)
T32
(P→R) ∧(Q→S) → (P∧Q → R∧S)
"Leibniz's praeclarum theorema"
T33
(P→Q) ∧(~P→Q) → Q
"separation of cases; constructive dilemma"
T34
(P→Q) ∧(P→~Q) → ~P
"reductio ad absurdum"
T35
(~P→R) ∧(Q→R) ↔ ((P→Q) → R)
T36
~(P ∧ ~P)
T37
(P→Q) ↔ ~(P ∧ ~Q)
Copyrighted material
"distribution of → over ∧"
"non-contradiction"
Chapter 2 -- 18
Version of Aug 2013
CHAPTER 2 SECTION 6
6 ABBREVIATING DERIVATIONS
It is useful in writing derivations to be able to combine two or more steps into one. For example, here is a
derivation in which double negation is used twice:
P
~Q → ~P
Q→R
∴ R
1. Show R
2. ~~P
3. ~~Q
4. Q
5. R
pr1 dn
2 pr2 mt
3 dn
4 pr3 mp dd
One can shorten this derivation by two steps by combining the double negations with other rules, like this:
1. Show R
2. ~~Q
3. R
pr1 dn pr2 mt
2 dn pr3 mp dd
The meanings of the notations at the end of the lines are:
2. " pr1 dn pr2 mt "
take pr1 and double negate it; then combine the result with pr2 by mt to get ~~Q
3. "2 dn pr3 mp "
double (un)negate the sentence on line 2; then combine the result with pr3 using
modus ponens
Here is a more highly abbreviated derivation.
P∧Q
R → ~Q
S∨~R → T
∴ T∧P
1. Show T ∧ P
2.
~R
pr1 s dn pr2 mt
simplify pr1 to get Q, then double negate Q to
get ~~Q; use mt on this and pr2 to get ~R
3.
T
2 add pr3 mp
apply add to the sentence on line 2 to get
S ∨ ~R; then apply mp to that and pr3 to get T
4.
T∧P
pr1 s 3 adj
simplify pr1 to get P and then adjoin this with the
sentence on line 3 to get T∧P.
dd
Abbreviations of this sort may always be interpreted by the following "decoding procedure", starting at the
left and moving right:
A line number or premise number gives you a sentence -- the sentence on that line.
Rule r also gives you a sentence -- the sentence on the line cited.
A sentence followed by 'dn', 's', 'add' or 'bc' gives you the result of applying that rule to that
sentence. (The old sentence is no longer available for further use.)
Two sentences followed by 'mp', 'mt', 'adj', 'cb' give you the result of applying that rule to them.
Copyrighted material
Chapter 2 -- 19
Version of Aug 2013
CHAPTER 2 SECTION 6
If you can apply this decoding and end up with the sentence on the line which has the abbreviations at its
end, the line is correct. If you can't, the line is not correct. (There is sometimes more than one way to
apply a rule to a sentence, so there may be many ways to use the decoding process. If at least one way
of using it ends you up with the sentence on the line, the abbreviation is correct; otherwise it is incorrect.)
Applied to the abbreviations on lines 2, 3 and 4 above the decoding looks like this. We work from the left.
First, the leftmost 'pr1' is replaced by the first premise:
2.
pr1
s
dn
pr2
mt
2.
P∧Q
s
dn
pr2
mt
dn
pr2
mt
Then rule s acts on this to produce 'Q':
2.
P∧Q
2.
s
Q
dn
pr2
mt
pr2
mt
pr2
mt
Then double negation turns 'Q' into '~~Q':
2.
Q
dn
2.
~~Q
Then pr2 is replaced by the second premise:
2.
~~Q
pr2
mt
2.
~~Q
R → ~Q
mt
Finally, rule mt acts on '~~Q' and 'R → ~Q' to give you '~R', which is the sentence that actually appears on
line 2:
2.
~~Q
pr2
2.
mt
~R
Line 3 is decoded by the same process:
3.
2
add
pr3
mp
3.
~R
add
pr3
mp
3.
S ∨ ~R
pr3
mp
3.
S ∨ ~R
S∨~R → T
mp
3.
Copyrighted material
T
Chapter 2 -- 20
Version of Aug 2013
CHAPTER 2 SECTION 6
Likewise for line 4:
4.
pr1
s
3
adj
4.
P∧Q
s
3
adj
4.
P
3
adj
4.
P
T
adj
4.
T∧P
A long string of abbreviations can be difficult to decode, so we will confine ourselves to simple cases.
EXERCISES
1. Use the method of abbreviating derivations to produce shortened derivations for T38, T40-43.
T38
P∧Q ↔ ~(P → ~Q)
T39
~(P∧Q) ↔ (P → ~Q)
T40
~(P → Q) ↔ P∧~Q
"negation of conditional"
T41
P ↔ P∧P
"idempotence for ∧"
T42
P∧~Q → ~(P→Q)
"negation of conditional"
T43
~P → ~(P∧Q)
T44
~Q → ~(P∧Q)
Copyrighted material
Chapter 2 -- 21
Version of Aug 2013
CHAPTER 2 SECTION 7
7 USING THEOREMS AS RULES
In Chapter 1 we learned a way to introduce instances of previously derived theorems into a derivation.
Theorems are even more useful when they are used to justify rules. The fundamental principle is that if a
theorem has been derived that has the form of a conditional, it can be cited as a rule which allows you to
infer one sentence from another whenever the conditional made from those sentences is an instance of
the theorem.
A theorem of conditional form:
∴ □→○
justifies a rule of this form:
□
∴ ○
For example, T13 ("transposition") is (P→Q) → (~Q→~P). This validates the rule:
□→○
∴ ~○ → ~□
We name such a rule by writing 'R' in front of the name of the theorem being used. An example of a use
of a theorem as a rule is:
8. S → T
9. ~T → ~S
8 RT13
Here are two arguments, and derivations, that use some theorems from Chapter 1 as rules.
~(Q∧~R) → P
P→Q
R → ~P
∴ ~(Q → R)
1. Show ~(Q → R)
2. Q → R
3. (Q→R) → (P→R)
4. P → R
5. (R→~P) → (P→~P)
6. P → ~P
7. ~P
8. ~~(Q∧~R)
9. Q ∧ ~R
10. Q
11. R
12. ~R
ass id
pr2 RT4
2 3 mp
4 RT4
pr3 5 mp
6 RT20
pr1 7 mt
8 dn
9s
2 10 mp
9 s 11 id
T4 is (P→Q) → ((Q→R) → (P→R))
T20 is (P→~P) → ~P
S→T
T → (Q → P)
S→Q
∴ S→P
1.
2.
3.
4.
Show S → P
S → (Q→P)
(S→Q) → (S→P)
S→P
Copyrighted material
pr1 pr2 RT4
2 RT6
pr3 3 mp dd
Chapter 2 -- 22
T4 is (P→Q) → ((Q→R) → (P→R))
T6 is (P→(Q→R)) → ((P→Q) → (P→R))
Version of Aug 2013
CHAPTER 2 SECTION 7
Theorems can be used to make rules in two more ways. One way applies when a theorem is a
biconditional. Since a biconditional is logically equivalent to two conditionals, it makes sense to use the
theorem as if it were two conditionals.
A theorem of biconditional form:
∴□↔○
validates both of these rules:
□
∴ ○
○
∴ □
An example is T27: '(P∧Q→R) ↔ (P→(Q→R))' which validates both of these:
□∧○ → △
∴ □ → (○→△)
RT27
RT27
□ → (○→△)
∴ □∧○ → △
A final additional way to use theorems as rules is possible when a theorem is a conditional whose
antecedent is a conjunction. This gives a rule which has multiple premises.
A theorem of this form:
∴ □∧○ → △
justifies a rule of this form:
□
○
∴ △
An example is T26:
(P→Q) ∧(Q→R) → (P→R)
which validates:
RT26
□→○
○→△
∴ □→△
These options combine, so that if one side of a biconditional is a conjunction, it validates a rule with
multiple premises. For example, T38 (below) is 'P∧Q ↔ ~(P→~Q)', and one of the rules that it validates
is:
RT38
□
○
∴ ~(□→~○)
Copyrighted material
Chapter 2 -- 23
Version of Aug 2013
CHAPTER 2 SECTION 7
EXERCISES
1. For each of the following derivations, determine which lines are correct and which incorrect. (In
assessing a line, assume that previous lines are correct.)
∴ ((U→V) → S) → (~S → U)
1. Show ((U→V) → S) → (~S → U)
2.
3.
(U→V) → S
Show ~S → U
ass cd
4.
5.
6.
7.
~S
~S → ~(U→V)
~(U → V)
U
ass cd
2 RT13
4 5 mp
6 RT21 cd
8.
T13 is (P→Q) → (~Q → ~P)
T21 is ~(P→Q) → P
3 cd
∴ ~V ∧ (W → V∧U) → (~W → ~S)
1. Show ~V ∧ (W → V∧U) → (~W → ~S)
2.
3.
4.
5.
6.
7.
8.
~V ∧ (W → V∧U)
~V
~(V∧U)
W → V∧U
~W
~(W∧S)
~W → ~S
ass cd
2s
3 RT43
2s
4 5 mt
6 RT43
7 RT39 cd
T43 is ~P → ~(P∧Q)
T39 is ~(P∧Q) ↔ (P→~Q)
Some more theorems:
T45 P∨Q ↔ (~P→Q)
T46 (P→Q) ↔ ~P ∨ Q
"definition of → in terms of ∨"
T47 P ↔ P∨P
"idempotence for ∨"
T48 (P∨Q) ∧(P→R) ∧(Q→S) → R∨S
T49 (P∨Q) ∧(P→R) ∧(Q→R) → R
"separation of cases"
T50 (P→R) ∧(Q→R) ↔ (P∨Q→R)
2. Construct a derivation for T45, and then use RT45 to derive T46
3. Construct derivations for T47 and T48, and construct a derivation for T49 using RT47 and RT48
4. Use RT49 in constructing a derivation for T50.
5. Derive T53.
Some additional theorems are given here for reference.
T51
(P∨Q) ∧(P→R) ∧(~P∧Q→R) → R
T52
(P→R) ∧(~P∧Q→R) ↔ (P∨Q→R)
T53
P∨Q ↔ Q∨P
Copyrighted material
"commutative law for ∨"
Chapter 2 -- 24
Version of Aug 2013
CHAPTER 2 SECTION 7
T54
P ∨ (Q∨R) ↔ (P∨Q) ∨ R
"associative law for ∨"
T55
(P → Q∨R) ↔ (P→Q) ∨ (P→R)
"distribution of → over ∨"
T56
(P→Q) → (R∨P → R∨Q)
T57
(P→Q) → (P∨R → Q∨R)
T58
(P→Q) ∨ (Q→R)
T59
P ∨ ~P
T60
(P→R) ∨ (Q→R) ↔ (P∧Q → R)
T61
P∧ (Q∨R) ↔ (P∧Q) ∨ (P∧R)
"distribution"
T62
P∨(Q∧R) ↔ (P∨Q) ∧(P∨R)
"distribution"
"excluded middle"
8 DERIVED RULES
We now have over fifty theorems that can be used as rules, with more to come. There are too many of
these to remember easily. It is customary to isolate a small number of rules based on the theorems and
give them special names, and use these rules frequently in derivations. In this section we look at five of
these.
Rule nc (negation of conditional): 'Negation of conditional' is an easy-to-remember name for rule
RT40 (which is "~(P→Q) ↔ P∧~Q"). It applies in either of these forms:
Rule nc
□ ∧ ~○
∴ ~(□→○)
~(□→○)
∴ □ ∧ ~○
This rule is often useful when you are trying to derive a conditional when a conditional derivation isn't
working for you. Instead of assuming the antecedent of the conditional in order to use cd, assume the
negation of the conditional for an indirect derivation. Then turn this negated conditional into a conjunction
of the antecedent with the negation of its consequent. This gives you a lot to work with in continuing the
derivation. As an example, suppose you are trying to validate this argument:
R→Q
R∨S
S→R
∴ P→Q
You begin the derivation:
1. Show P → Q
You may now assume 'P' for purposes of doing a conditional derivation. But 'P' does not occur among the
premises, and you may not see how to proceed. Instead of trying a conditional derivation, begin an
indirect derivation:
1. Show P → Q
2.
~(P→Q)
ass id
Then apply the derived rule nc:
Copyrighted material
Chapter 2 -- 25
Version of Aug 2013
CHAPTER 2 SECTION 8
1. Show P → Q
2. ~(P→Q)
3. P ∧ ~Q
ass id
2 nc
It is now fairly easy to simplify off '~Q', and use it to derive a contradiction:
1. Show P → Q
2. ~(P→Q)
3. P ∧ ~Q
4. ~Q
5. ~R
6. S
7. R
ass id
2 nc
3s
4 pr1 mt
5 pr2 mtp
6 pr3 mp

contradictories

So you can finish the indirect derivation:
1. Show P → Q
2.
3.
4.
5.
6.
7.
~(P→Q)
P ∧ ~Q
~Q
~R
S
R
ass id
2 nc
3s
4 pr1 mt
5 pr2 mtp
6 pr3 mp 5 id
Rule cdj (conditional as disjunction): This rule is constituted by T45 and T46, which together assert
the equivalence of a conditional with a disjunction whose left disjunct is the negation (or unnegation) of the
antecedent of the conditional and whose right disjunct is the consequent. Rule cdj has four cases:
Rule cdj
□→○
∴ ~□ ∨ ○
~□ ∨ ○
∴ □→○
~□ → ○
∴ □∨○
□∨○
∴ ~□ → ○
This rule can be useful when attempting to derive a disjunction. Instead of deriving the disjunction directly,
derive the conditional whose antecedent is the (un)negation of the left disjunct and whose consequent is
the other disjunct. This can usually be done using a conditional derivation. Then turn the result of the
conditional derivation into the disjunction you are after using derived rule cdj.
Here is a derivation of T54 using cdj together with T53, which was derived in the exercises for the last
section. The overall structure of the derivation is to derive the biconditional by using two conditional
derivations to get the corresponding conditionals, and then use rule cb.
Copyrighted material
Chapter 2 -- 26
Version of Aug 2013
CHAPTER 2 SECTION 8
T54
P ∨ (Q∨R) ↔ (P∨Q) ∨ R
1. Show P ∨ (Q∨R) ↔ (P∨Q) ∨ R
Show P ∨ (Q∨R) → (P∨Q) ∨ R
2.
3.
4.
P ∨ (Q∨R)
Show ~R → (P∨Q)
ass cd
5.
6.
~R
Show ~P → Q
ass cd
7.
8.
9.
~P
Q∨R
Q
10.
P∨Q
11.
12.
R ∨ (P∨Q)
(P∨Q) ∨ R
ass cd
3 7 mtp
5 8 mtp cd
6 cdj cd
 cdj
4 cdj
11 RT53 cd
 cdj
13. Show (P∨Q) ∨ R → P ∨ (Q∨R)
24.
LIKE LINES 3-12
<with two uses of cdj>
P ∨ (Q∨R) ↔ (P∨Q) ∨ R
2 13 cb
 cdj twice
Rule sc (separation of cases): This is a combination of RT33 and RT49. It validates these inferences:
Rule sc
□∨○
□→△
○→△
∴ △
□→△
~□ → △
∴△
The first form of rule sc (on the left) says that if you are given that at least one of two cases hold (the first
premise), and if each of them imply something (the second and third premises), then you can conclude
that thing.
The second form of rule sc (on the right) applies when one of the two cases is the negation of the other.
Then their disjunction (P ∨ ~P) is logically true, and needn't be stated as an additional premise. (See
below for illustration.)
Rule sc is especially useful when other attempts to produce a derivation have failed. For example, if you
have a disjunction on an available line, then see if you can do two conditional derivations, each starting
with one of the disjuncts, and each reasoning to the desired conclusion. If you can do this, the first form of
sc applies. As an example, suppose you are given this argument:
V∨W
W → ~X
~U → X
∴ U∨V
It may not be apparent how to proceed. So consider separation of cases. You have available a
disjunction, 'V ∨ W', which is the first premise. If you can derive both V → U∨V and W → U∨V, the rule sc
will give you the desired conclusion:
Copyrighted material
Chapter 2 -- 27
Version of Aug 2013
CHAPTER 2 SECTION 8
1. Show U ∨ V
2.
Show V → U∨V
3.
4.
V
U∨V
5.
Show W → U∨V
6.
7.
8.
9.
10.
W
~X
U
U∨V
U∨V
ass cd
3 add cd
ass cd
6 pr2 mp
7 pr3 mt dn
8 add cd
pr1 2 5 sc dd
When you don't have a disjunction to work with, you may be able to use the second form of sc. Suppose
you have this argument:
R∧S → Q
R→S
∴ R→Q
In applying the second form of sc, you need to choose something which will serve as the antecedent for a
conditional whose consequent is the desired conclusion, and whose negation will also serve as the
antecedent for a conditional whose consequent is the desired conclusion. What should you choose?
Often there is more than one choice that will work. In the case we are given, 'R' will work for this purpose.
That is, you will indeed be able to derive both of these:
R → (R→Q)
~R → (R→Q)
The second form of rule sc will then give you the desired conclusion:
1. Show R → Q
2.
3.
4.
5.
Show R → (R→Q)
R
S
Q
 derive 'R → (R→Q)'
ass cd
3 pr2 mp
3 4 adj pr1 mp cd
6.
Show ~R → (R → Q)
7.
8.
~R
Show R → Q
9.
10.
R
~R
11.
12.
 derive '~R → (R→Q)'
ass cd
ass cd
7 r id
8 cd
R→Q
2 6 sc
dd
 apply the second form of sc
Rule dm (DeMorgan's): This is a very useful rule. It lets you replace negations of conjunctions with
modified disjunctions, and vice versa. It consists of any application of the rules based on theorems T63T66:
Copyrighted material
Chapter 2 -- 28
Version of Aug 2013
CHAPTER 2 SECTION 8
T63
P ∧ Q ↔ ~(~P∨~Q)
T64
P ∨ Q ↔ ~(~P∧~Q)
T65
~(P∧Q) ↔ ~P ∨ ~Q
T66
~(P∨Q) ↔ ~P ∧ ~Q
So it allows any of the following inferences:
Rule dm
□∧○
∴ ~(~□∨~○)
∴
~(~□∨~○)
□∧○
□∨○
∴ ~(~□∧~○)
~(□∧○)
∴ ~□ ∨ ~○
~(□∨○)
∴ ~□ ∧ ~○
~(~□∧~○)
□∨○
~□ ∨ ~○
∴ ~(□∧○)
~□ ∧ ~○
∴ ~(□∨○)
∴
It may be easiest to remember these forms by remembering T63 and T64 in this form:
A negation of a
conjunction
disjunction
is equivalent to the
disjunction
of the negations of its parts.
conjunction
DeMorgan's rule can be handy when you are trying to derive a disjunction. To use it, you assume the
negation of the disjunction for an indirect derivation. Rule dm lets you turn that negation into a
conjunction, and then you have both conjuncts to use in deriving a contradiction. Example:
P→U
P∨Q
Q→V
∴ U∨V
1. Show U ∨ V
2.
3.
4.
5.
6.
7.
~(U∨V)
~U ∧ ~V
~P
Q
V
~V
ass id
2 dm
3 s pr1 mt
4 pr2 mtp
5 pr3 mp
3 s 6 id
Rule nb (negation of biconditional): This rule is an application of T90:
~(P↔Q) ↔ (P↔~Q)
The rule sanctions these inferences:
Rule nb
~(□↔○)
∴ □ ↔ ~○
Copyrighted material
□ ↔ ~○
∴ ~(□↔○)
Chapter 2 -- 29
Version of Aug 2013
CHAPTER 2 SECTION 8
The first form is handy if you have the negation of a biconditional. The rule lets you infer a biconditional,
which simplifies into two conditionals, which can be very useful. Here is an example:
~(P↔Q)
~Q
∴ P
1. Show P
2.
3.
4.
5.
~(P ↔ Q)
P ↔ ~Q
~Q → P
P
pr
2 nb
3 bc
4 pr2 mp
dd
The second form is handy if you want to derive the negation of a biconditional. Just derive the related
biconditional, say by using conditional derivations to derive the associated conditionals. Example:
P → (R↔Q)
R → ~Q
S→Q
~R → S
∴ ~P
1. Show ~P
2.
Show ~Q → R
3.
4.
5.
6.
~Q
~S
~~R
R
ass cd
3 pr3 mt
4 pr4 mt
5 dn cd
7.
8.
9.
R ↔ ~Q
~(R↔Q)
~P
pr2 2 cb
7 nb
8 pr1 mt
dd
EXERCISES
1. For each of the following derivations, determine which lines are correct and which incorrect. (In
assessing a line, assume that previous lines are correct.)
a.
∴
(U→S) → Q
P∨R → S
~(T→Q)
~P
1. Show ~P
2.
3.
4.
5.
6.
7.
T ∧ ~Q
~(U→S)
U ∧ ~S
~(P∨R)
~P ∧ ~R
~P
Copyrighted material
pr3 nc
2 s pr1 mt
3 nc
4 s pr2 mt
5 dm
6 s dd
Chapter 2 -- 30
Version of Aug 2013
CHAPTER 2 SECTION 8
b.
~X ∨ W
~(V ↔ W)
~(W ↔X) ∨ V
∴ ~W
1. Show ~W
2.
3.
4.
c.
Show W → X
W
X
ass cd
3 pr1 mtp
5.
Show X → W
6.
7.
X
X→W
8.
9.
10.
11.
W↔X
V
V ↔ ~W
~W
ass cd
pr1 cdj dd
2 5 bc
8 dn pr3 mtp
pr2 nb
9 10 mp dd
(X →U) → (Y→Z)
~(Y ∨ ~Z)
∴ ~U
1. Show ~U
2.
3.
4.
5.
6.
~Y ∧ Z
~(Y → Z)
~(X → U)
X ∧ ~U
~U
pr2 dm
2 nc
3 4 mt
4 nc
5 s dd
2. Construct correct derivations for each of the following arguments using derived rules when convenient.
a.
U∧V → X
~V → Y
X∨Y → Z
∴ ~Z → ~U
<use dm>
b.
(X→Y) → Z
~Z
V→Y
∴ ~V
<use nc>
c.
P∨Q
Q→S
U ∨ ~S
P∨S → R
R→U
∴ U
Copyrighted material
Chapter 2 -- 31
Version of Aug 2013
CHAPTER 2 SECTION 9
9 OFFICIAL CONDITIONS FOR DERIVATIONS
Let us summarize here what we can now use in constructing an unabbreviated derivation.
UNABBREVIATED DERIVATIONS
A derivation from a set of sentences P consists of a sequence of lines that is built up in order, step by
step, where each step is in accordance with these provisions:
•
•
•
•
•
•
•
•
•
Show line: A show line consists of the word "Show" followed by a symbolic sentence. A show
line may be introduced at any step. Show lines are not given a justification.
Premise: At any step, any symbolic sentence from the set P may be introduced, justified with the
notation "pr".
Theorem: At any step, an instance of a previously proved theorem may be entered with the name
of the theorem given as justification. (e.g. "T32")
Rule: At any step, a line may be introduced if it follows by a rule from sentences on previous
available lines; it is justified by citing the numbers of those previous lines and the name of the rule.
This includes the following basic rules:
r
mp
mt
dn
s
adj
add
mtp
bc
cb
It also includes rules based on previously derived theorems, where the name of a rule based on a
theorem is "R" followed by the name of the theorem; e.g. "RT32". If the appropriate enabling
theorems have been derived, these rules are also available for use:
nc
cdj
sc
dm
nb
Direct derivation: When a line (which is not a show line) is introduced whose sentence is the
same as the sentence on the closest previous uncancelled show line, one may, as the next step,
write "dd" following the justification for that line, draw a line through the word "Show", and draw a
box around all the lines below the show line, including the current line.
Assumption for conditional derivation: When a show line with a conditional sentence is
introduced, as the next step one may introduce an immediately following line with the antecedent
of the conditional on it; the justification is "ass cd".
Conditional derivation: When a line (which is not a show line) is introduced whose sentence is
the same as the consequent of the conditional sentence on the closest previous uncancelled
show line, one may, as the next step, write "cd" at the end of that line, draw a line through the
word "Show", and draw a box around all the lines below the show line, including the current line.
Assumption for indirect derivation: When a show line is introduced, as the next step one may
introduce an immediately following line with the [un]negation of the sentence on the show line; the
justification is "ass id".
Indirect derivation: When a sentence is introduced on a line which is not a show line, if there is a
previous available line containing the [un]negation of that sentence, and if there is no uncancelled
show line between the two sentences, as the next step you may write the line number of the first
sentence followed by "id" at the end of the line with the second sentence. Then you cancel the
closest previous "show", and box all sentences below that show line, including the current line.
Except for steps that involve boxing and canceling, every step introduces a line. When writing out a
derivation, every line that is introduced is written directly below previously introduced lines.
Optional variant: When boxing and canceling with direct or conditional derivation, the "dd" or "cd"
justification may be written on a later line which contains no sentence at all, and which is followed by the
number of the line that completes the derivation. With indirect derivation, the "id" justification may be
written on a later line which contains no sentence at all, and which is followed by the numbers of the two
lines containing contradictory sentences. In all cases, the lines cited must be available from the later line.
Copyrighted material
Chapter 2 -- 32
Version of Aug 2013
CHAPTER 2 SECTION 9
Some additional strategic hints
Now that we have connectives in addition to the negation and conditional signs, we can give some general
hints for doing derivations containing them. These have all been illustrated above, and they will simply be
stated here for convenience. First are strategies that are often useful for deriving certain forms of
sentences.
If you want to derive a Conjunction □ ∧ ○
Derive each conjunct (perhaps by id) and adjoin them
If you want to derive a Disjunction □ ∨ ○
Derive either disjunct and use add.
Assume '~(□ ∨ ○)' for id, and use dm.
Derive '~□ → ○', perhaps by cd, and use cdj
If you want to derive a Biconditional □ ↔ ○
Derive each conditional and use cb.
If you want to derive a Negation of a conjunction ~(□ ∧ ○)
Use id.
If you want to derive a Negation of a disjunction ~(□ ∨ ○)
Derive '~□ ∧ ~○' and use dm.
Perhaps assume '□ ∨ ○' for id, and try to derive both '□ → P∧~P' and '○→P∧~P'. Then use sc
(applied to the assumed '□ ∨ ○' and the conditionals) to derive 'P∧~P'.
If you want to derive a Negation of a biconditional ~(□ ↔ ○)
Derive '□ ↔ ~○' and use nb.
Then there are situations in which you have available a certain form of sentence, and want to know how to
make use of it.
If you have available a Conjunction □ ∧ ○
Simplify and use the conjuncts singly.
If you have available a Disjunction □ ∨ ○
Try to derive the negation of one of the disjuncts, and use mtp
Derive the conditionals '□ → △' and '○ → △', where '△' is something you want to derive. Then
use sc with the disjunction and two conditionals.
If you have available a Biconditional □ ↔ ○
Infer both conditionals and use them with mp, mt, and so on.
If you have available a Negation of a conjunction ~(□ ∧ ○)
Use dm to turn this into '~□ ∨ ~○', and then try to derive either '□' or '○' to use mtp.
If you have available a Negation of a disjunction ~(□ ∨ ○)
Use dm to turn this into '~□ ∧ ~○'; then simplify and use the conjuncts singly.
If you have available a Negation of a biconditional ~(□ ↔ ○)
Use nb to turn this into '□ ↔ ~○', and use bc to get the corresponding conditionals.
Copyrighted material
Chapter 2 -- 33
Version of Aug 2013
CHAPTER 2 SECTION 9
EXERCISES
1. Construct correct derivations for each of the following arguments.
a.
~(P ↔ Q)
R∨P
~Q → R
∴ R
<use nb>
b.
W→U
~W → V
∴ U∨V
<use cdj>
c.
P ∨ (Q∧S)
R∨Q
S ∨ ~P
Q → ~S
∴ R
10 TRUTH TABLES AND TAUTOLOGIES
The sentence 'P ∨ ~P' is logically true. It is true in all logically possible situations. This can be established
by simple reasoning:
Although there are an infinite number of logically possible situations, they fall into two classes.
One class consists of situations in which 'P' is true, and the other consists of situations in which 'P'
isn't true. In any situation in the first class, 'P ∨ ~P' is true because it is a disjunction whose first
disjunct is true. In any situation in the second class, 'P' is not true in that situation, and so its
negation, '~P' is true, and again 'P ∨ ~P' is true because it is a disjunction which has a true
disjunct. So in either class of situations 'P ∨ ~P' is true.
This pattern of reasoning can be summed up using a truth table. The table begins with listing the two
options for the truth value of 'P' in a class of situations:
situations in which 'P' is true 
situations in which 'P' is false 
P
T
F
~P
The truth value of '~P' is determined in each class:
situations in which 'P' is true 
situations in which 'P' is false 
P
T
F
~P
F
T
and that information determines the truth value of 'P ∨ ~P' in each class:
situations in which 'P' is true 
situations in which 'P' is false 
P
T
F
~P
F
T
P ∨ ~P
T
T
This is an example in which no matter what truth value the simple parts of the sentence have, the
sentence itself is true. Such a sentence is called a "tautology":
Copyrighted material
Chapter 2 -- 34
Version of Aug 2013
CHAPTER 2 SECTION 10
Tautology: A sentence is a tautology if no matter what truth values are
assigned to its simple parts, the definitions of the connectives used in the
sentence determine that the sentence is true.
The truth table just given shows that 'P ∨ ~P' is a tautology, because it shows that 'P ∨ ~P' is true no
matter how truth values are assigned to its atomic parts.
This use of truth tables can be applied to a sentence of any degree of complexity. Here is an example
showing that 'P∧Q → Q' is a tautology. We begin by listing all of the possible combinations of truth values
that 'P' and 'Q' might have. There are four of these: the sentences are both true, the first is true and the
second false, the first is false and the second true, or they are both false:
P
T
T
F
F
Q
T
F
T
F
P∧Q
P∧Q → Q
This assignment of truth values to 'P' and to 'Q' determines the truth values of 'P ∧ Q' and of 'P∧Q → Q';
for 'P∧Q → Q':
P
Q
P∧Q
P∧Q → Q
T
T
T
T
T
F
F
T
F
T
F
T
F
F
F
T
In this table there are all T's under 'P∧Q → Q', showing that it is a tautology.
To handle sentences of arbitrary numbers of sentence letters, we need to have a systematic way of
representing all of the possible combinations of truth values that the sentence letters can receive. One
way of doing this is to list all of the atomic parts of the sentence on the top of the table. Then, underneath
the rightmost letter, write alternations of T and F:
P
Q
R
T
F
T
F
T
F
T
F
Under the next letter to its left write alterations of TT and FF:
P
Q
T
T
F
F
T
T
F
F
R
T
F
T
F
T
F
T
F
Under the next, write alterations of TTTT and FFFF:
Copyrighted material
Chapter 2 -- 35
Version of Aug 2013
CHAPTER 2 SECTION 10
P
T
T
T
T
F
F
F
F
Q
T
T
F
F
T
T
F
F
R
T
F
T
F
T
F
T
F
Do this until the leftmost letter has gone through one whole set of alterations. If there is one sentence
letter, only two rows are required. If there are two, the table will contain four rows. If three, then eight.
n
And so on. There are always 2 rows in the table when there are n sentence letters.
Next, write the sentence to be tested, and underneath it write in each row the truth value that it has when
its parts have the truth values appearing on that row. An example is:
P
T
T
T
T
F
F
F
F
Q
T
T
F
F
T
T
F
F
R
T
F
T
F
T
F
T
F
P∧Q → P∧R
T
F
T
T
T
T
T
T
 not a tautology
Since there is a row (the second row) in which 'P∧Q → P∧R' does not have a T, that sentence is not a
tautology.
If there is a T in every row, it is a tautology, as in this case:
P
T
T
T
T
F
F
F
F
Q
T
T
F
F
T
T
F
F
R
T
F
T
F
T
F
T
F
P∧Q → P∨R
T
T
T
T
T
T
T
T
This method is completely mechanical and it always yields an answer in a finite amount of time.
If a sentence is a tautology, then you need to fill in every one of its rows to show that every one is T. But if
a sentence is not a tautology, you need only find one row in which the sentence comes out F. For
example, the following partial truth table shows that 'P∨Q ↔ R∨Q' is not a tautology:
Copyrighted material
Chapter 2 -- 36
Version of Aug 2013
CHAPTER 2 SECTION 10
P
T
T
T
T
F
F
F
F
Q
T
T
F
F
T
T
F
F
R
T
F
T
F
T
F
T
F
P∨Q ↔ R∨Q
F
If a sentence is not a tautology and if you can identify what assignment of truth values to the simple parts
will show this, you needn't set up a whole truth table. Just give a single row:
P
T
Q
F
R
F
P∨Q ↔ R∨Q
F
and state that this assignment of truth values to sentence letters makes the sentence false.
It turns out that every theorem of the first two chapters of this text is a tautology. This is because the rules
and techniques in these chapters only allow derivations of tautologies when no premises are used. It is
also a fact that any tautology can be derived by the rules we have. So we have two ways to show
tautologies: theorems and truth tables, and we have one way to show non-tautologies: truth tables.
EXERCISES
1. Use truth tables or truth value assignments to determine whether each of these is a tautology.
a.
b.
c.
d.
e.
f.
(R↔S) ∨ (R↔~S)
R ↔ (S↔R)
R ∨ (S∧T) → R ∧ (S∨T)
~U → (U→~V)
(~R↔R) → S
(S∧T) ∨ (S∧~T) ∨ ~S
11 TAUTOLOGICAL VALIDITY
It is easy to show by doing a derivation that this argument is valid:
P∧P
∴ P
It is also possible to show that the argument is valid using a technique like that of truth tables. Just show
that there is no logically possible situation in which the premise is true and the conclusion false. This can
be done as follows:
All logically possible situations can be divided into two classes. In one class of situations, 'P' is
true; no situation of this sort can be one in which the argument has true premises and a false
conclusion, because in any of these situations the conclusion, 'P', is true. In all other situations,
'P' is false. But then so is 'P∧P'. So none of these are situations in which the argument has true
premises and a false conclusion. So it is valid.
Generalizing, we can say that an argument is valid whenever the premises "tautologically imply" the
conclusion. This relation is called Tautological Validity. It is defined as follows:
Copyrighted material
Chapter 2 -- 37
Version of Aug 2013
CHAPTER 2 SECTION 11
An argument is tautologically valid if and only if there is no assignment of
truth values to its atomic parts which make the premises all true and the
conclusion false.
There is a mechanical way to test an argument to see if it is tautologically valid. Just create a truth table in
which all of the premises and conclusion appear at the top of some column. If there is no row in which all
of the premises have T's under them and the conclusion has an F, then the argument is tautologically
valid. If there is such a row, then the argument is not tautologically valid; instead, we say that it is
tautologically invalid.
Suppose that we are wondering whether this argument:
P → ~Q
R ↔ P∧Q
Q∨R
∴ Q∧R
is tautologically valid. Here is a truth table to test this:
P
T
T
T
T
F
F
F
F
Q
T
T
F
F
T
T
F
F
R
T
F
T
F
T
F
T
F
P → ~Q
F
F
T
T
T
T
T
T
R ↔ P∧Q
T
F
F
T
F
T
F
T
Q∨R
T
T
T
F
T
T
T
F
Q∧R
T
F
F
F
T
F
F
F
There is in fact a row (the sixth) in which the premises are all true and the conclusion is false. So the
argument is not tautologically valid.
Using the same technique, we can show that this argument:
∴
P → ~Q
R ↔ P∧Q
Q∨R
P↔R
is tautologically valid. The truth table is:
P
T
T
T
T
F
F
F
F
Q
T
T
F
F
T
T
F
F
R
T
F
T
F
T
F
T
F
P → ~Q
F
F
T
T
T
T
T
T
R ↔ P∧Q
T
F
F
T
F
T
F
T
Q∨R
T
T
T
F
T
T
T
F
P↔R
T
F
T
F
F
T
F
T
Here, there is no row in which the premises are all true and the conclusion 'P ↔ R' is false. So that
argument is tautologically valid.
Although we haven’t shown this here, if there is a derivation using the rules and techniques of Chapters 1
and 2 showing that an argument is valid, then the argument is indeed tautologically valid. And vice versa:
Copyrighted material
Chapter 2 -- 38
Version of Aug 2013
CHAPTER 2 SECTION 11
if an argument is tautologically valid, then there is a derivation to show that the argument is valid using the
techniques of chapters 1 and 2.
So we have two ways to show that an argument is tautologically valid: it can be done with a truth table or
with a derivation. We have only one way to show that some argument is not tautologically vaild: find a way
to assign truth values to the sentential letters so that all the premises are true and the conclusion false.
EXERCISES
For each of the following arguments, either show that it is tautologically valid, or show that it is
tautologically invalid.
a.
U∧V → X
~V → U
X∨V → U
∴ V → ~U
b.
(X→Y) → Z
~Z
∴ ~Y
c.
~(P ↔ Q)
R∨P
~Q → R
∴ R
d.
S∨T
W∨S
~T ∨ ~S
∴ ~S
e.
W→U
~W → V
∴ U∨V
f.
P ↔ ~Q
Q → R∨P
R → ~Q ∨ ~P
∴ Q∨R
g.
P ∨ (Q∧S)
S∨Q
S ∨ ~P
∴ S
h.
P ∧ (Q∨S)
S∨Q
S∨P
∴ S
Copyrighted material
Chapter 2 -- 39
Version of Aug 2013
CHAPTER 2 SUMMARIES
BASIC AND DERIVED RULES FROM CHAPTERS 1 AND 2
BASIC RULES OF CHAPTER 1
Repetition:
□
∴ □
Modus Ponens:
□→○
□
∴○
Modus Tollens:
□→○
~○
∴ ~□
Double negation:
□
∴ ~~□
or
~~□
∴□
BASIC RULES OF CHAPTER 2
Conjunction rules:
Rule s (simplification)
□∧○
∴ □
or
Rule adj (adjunction)
□∧○
∴○
Disjunction rules:
Rule add (addition)
□
∴ □∨○
or
Rule mtp (modus tollendo ponens)
□
∴○∨□
Biconditional rules:
Rule bc (biconditional-to-conditional)
□↔○
∴ □→○
Copyrighted material
or
□
○
∴ □∧○
□↔○
∴○→□
Chapter 2 -- 40
□∨○
~○
∴ □
or
□∨○
~□
∴○
Rule cb (conditionals-to-biconditional)
□→○
○→□
∴ □↔○
Version of Aug 2013
CHAPTER 2 SUMMARIES
THEOREMS USED AS RULES
A theorem of conditional form:
∴ □∧○ → △
justifies a rule of this form:
□
○
∴ △
A theorem of biconditional form:
∴□↔○
justifies whatever both conditionals justify.
DERIVED RULES (May be used if the theorems on which they are based have been
derived.)
Rule nc
<based on T40>
□ ∧ ~○
∴ ~(□→○)
~(□→○)
∴ □ ∧ ~○
Rule cdj
<based on T45, T46>
□→○
∴ ~□ ∨ ○
Rule sc
~□ ∨ ○
∴ □→○
~□ → ○
∴ □∨○
□∨○
∴ ~□ → ○
<based on T33, T49>
□∨○
□→△
○→△
∴ △
Rule dm
<based on T63-66>
□∧○
∴ ~(~□∨~○)
∴
Rule nb
~(~□∨~○)
□∧○
□∨○
∴ ~(~□∧~○)
~(□∧○)
∴ ~□ ∨ ~○
~(□∨○)
∴ ~□ ∧ ~○
~(~□∧~○)
□∨○
~□ ∨ ~○
∴ ~(□∧○)
~□ ∧ ~○
∴ ~(□∨○)
∴
<based on T90>
~(□↔○)
∴ □ ↔ ~○
Copyrighted material
□→△
~□ → △
∴△
□ ↔ ~○
∴ ~(□↔○)
Chapter 2 -- 41
Version of Aug 2013
CHAPTER 2 SUMMARIES
UNABBREVIATED DERIVATIONS
A derivation from a set of sentences P consists of a sequence of lines that is built up in order, step by
step, where each step is in accordance with these provisions:
•
•
•
•
•
•
•
•
•
Show line: A show line consists of the word "Show" followed by a symbolic sentence. A show
line may be introduced at any step. Show lines are not given a justification.
Premise: At any step, any symbolic sentence from the set P may be introduced, justified with the
notation "pr".
Theorem: At any step, an instance of a previously proved theorem may be entered with the name
of the theorem given as justification. (e.g. "T32")
Rule: At any step, a line may be introduced if it follows by a rule from sentences on previous
available lines; it is justified by citing the numbers of those previous lines and the name of the rule.
This includes the following basic rules:
r
mp
mt
dn
s
adj
add
mtp
bc
cb
It also includes rules based on previously derived theorems, where the name of a rule based on a
theorem is "R" followed by the name of the theorem; e.g. "RT32". If the appropriate enabling
theorems have been derived, these rules are also available for use:
nc
cdj
sc
dm
nb
Direct derivation: When a line (which is not a show line) is introduced whose sentence is the
same as the sentence on the closest previous uncancelled show line, one may, as the next step,
write "dd" following the justification for that line, draw a line through the word "Show", and draw a
box around all the lines below the show line, including the current line.
Assumption for conditional derivation: When a show line with a conditional sentence is
introduced, as the next step one may introduce an immediately following line with the antecedent
of the conditional on it; the justification is "ass cd".
Conditional derivation: When a line (which is not a show line) is introduced whose sentence is
the same as the consequent of the conditional sentence on the closest previous uncancelled
show line, one may, as the next step, write "cd" at the end of that line, draw a line through the
word "Show", and draw a box around all the lines below the show line, including the current line.
Assumption for indirect derivation: When a show line is introduced, as the next step one may
introduce an immediately following line with the [un]negation of the sentence on the show line; the
justification is "ass id".
Indirect derivation: When a sentence is introduced on a line which is not a show line, if there is a
previous available line containing the [un]negation of that sentence, and if there is no uncancelled
show line between the two sentences, as the next step you may write the line number of the first
sentence followed by "id" at the end of the line with the second sentence. Then you cancel the
closest previous "show", and box all sentences below that show line, including the current line.
Except for steps that involve boxing and canceling, every step introduces a line. When writing out a
derivation, every line that is introduced is written directly below previously introduced lines.
Optional variant: When boxing and canceling with direct or conditional derivation, the "dd" or "cd"
justification may be written on a later line which contains no sentence at all, and which is followed by the
number of the line that satisfies the conditions for direct or conditional derivation. With indirect derivation,
the "id" justification may be written on a later line which contains no sentence at all, and which is followed
by the numbers of the two lines containing contradictory sentences. In all cases, the lines cited must be
available from the later line.
Copyrighted material
Chapter 2 -- 42
Version of Aug 2013
CHAPTER 2 SUMMARIES
STRATEGY HINTS
Try to reason out the argument for yourself.
Begin with a sketch of an outline of a derivation, and then fill in the details.
Write down obvious consequences.
When no other strategy is obvious, try indirect derivation.
To derive:
Try this:
Conjunction
□∧○
Derive each conjunct, and adjoin them
Disjunction
□∨○
Derive either disjunct and use add. (Often this is not possible.)
Assume '~(□ ∨ ○)' for id and immediately use dm.
Derive '~□ → ○' and use cdj
Conditional
□→○
Use cd
Biconditional
□↔○
Derive both conditionals and use cb.
Negation of
conjunction
~(□ ∧ ○)
Use id.
Negation of
disjunction
~(□ ∨ ○)
Derive '~□ ∧ ~○' and use dm.
Perhaps assume '□ ∨ ○' for id and try to derive both '□ → P∧~P' and '○→P∧~P'.
Then use sc (applied to the assumed '□ ∨ ○' and the conditionals) to derive 'P∧~P'.
Negation of
conditional
~(□ → ○)
Use id.
Negation of
biconditional
~(□ ↔ ○)
Derive '□ ↔ ~○' and use nb.
If you have this
available:
Try this:
Conjunction
□∧○
Simplify and use the conjuncts singly.
Disjunction
□∨○
Try to derive the negation of one of the disjuncts, and use mtp..
Derive the conditionals '□ → △' and '○ → △', where '△' is something you want to
derive. Then use sc with the disjunction and two conditionals.
Conditional
□→○
Try to derive the antecedent to set up mp, or derive the negation of the consequent,
to set up mt.
Biconditional
□↔○
Infer both conditionals and use them with mp, mt, and so on.
Copyrighted material
Chapter 2 -- 43
Version of Aug 2013
CHAPTER 2 SUMMARIES
Negation of
conjunction
~(□ ∧ ○)
Use dm to turn this into '~□ ∨ ~○', and then try to derive either '□' or '○' to use mtp.
Negation of
disjunction
~(□ ∨ ○)
Use dm to turn this into '~□ ∧ ~○'; then simplify and use the conjuncts singly.
Negation of
conditional
~(□ → ○)
Use nc to derive '□ ∧ ~○', then simplify and use the conjuncts singly.
Negation of
biconditional
~(□ ↔ ○)
Use nb to turn this into '□ ↔ ~○', and use bc to get the corresponding conditionals.
ALL THEOREMS FROM CHAPTERS 1 AND 2
T1
P→P
T2
Q → (P→Q)
T3
P → ((P→Q) → Q)
T4
(P→Q) → ((Q→R) → (P→R))
Syllogism
T5
(Q→R) → ((P→Q) → (P→R))
Syllogism
T6
(P → (Q→R)) → ((P→Q) → (P→R))
Distribution of → over →
T7
((P→Q) → (P→R)) → (P→(Q → R))
Distribution of → over →
T8
(P→ (Q→R)) → (Q → (P→R))
Commutation
T9
(P → (P→Q)) → (P→Q)
T10
((P→Q) → Q) → ((Q→P) → P)
T11
~~P → P
Double negation
T12
P → ~~P
Double negation
T13
(P→Q) → (~Q → ~P)
Transposition
T14
(P → ~Q) → (Q → ~P)
Transposition
T15
(~P → Q) → (~Q → P)
Transposition
T16
(~P → ~Q) → (Q → P)
Transposition
T17
P → (~P→Q)
T18
~P → (P→Q)
T19
(~P→P) → P
Reductio ad absurdum
T20
(P→~P) → ~P
Reductio ad absurdum
T21
~(P→Q) → P
T22
~(P→Q) → ~Q
T23
((P→Q) → P) → P
Copyrighted material
Peirce's law
Chapter 2 -- 44
Version of Aug 2013
CHAPTER 2 SUMMARIES
T24
P∧Q ↔ Q∧P
"commutative law for conjunction"
T25
P∧ (Q∧R) ↔ (P∧Q)∧ R
"associative law for conjunction"
T26
(P→Q) ∧(Q→R) → (P→ R)
"hypothetical syllogism"
T27
(P∧Q → R) ↔ (P → (Q→R))
"exportation"
T28
(P∧Q → R) ↔ (P∧~R → ~Q)
T29
(P→ Q∧R) ↔ (P→Q) ∧(P→R)
T30
(P→Q) → (R∧P → R∧Q)
T31
(P→Q) → (P∧R → Q∧R)
T32
(P→R) ∧(Q→S) → (P∧Q → R∧S)
"Leibniz's praeclarum theorema"
T33
(P→Q) ∧(~P→Q) → Q
"separation of cases; constructive dilemma"
T34
(P→Q) ∧(P→~Q) → ~P
"reductio ad absurdum"
T35
(~P→R) ∧(Q→R) ↔ ((P→Q) → R)
T36
~(P ∧ ~P)
T37
(P→Q) ↔ ~(P ∧ ~Q)
T38
P∧Q ↔ ~(P → ~Q)
T39
~(P∧Q) ↔ (P → ~Q)
T40
~(P → Q) ↔ P∧~Q
"negation of conditional"
T41
P ↔ P∧P
"idempotence for ∧"
T42
P∧~Q → ~(P→Q)
"negation of conditional"
T43
~P → ~(P∧Q)
T44
~Q → ~(P∧Q)
T45
P∨Q ↔ (~P→Q)
T46
(P→Q) ↔ ~P ∨ Q
"definition of → in terms of ∨"
T47
P ↔ P∨P
"idempotence for ∨"
T48
(P∨Q) ∧(P→R) ∧(Q→S) → R∨S
T49
(P∨Q) ∧(P→R) ∧(Q→R) → R
T50
(P→R) ∧(Q→R) ↔ (P∨Q→R)
T51
(P∨Q) ∧(P→R) ∧(~P∧Q→R) → R
T52
(P→R) ∧(~P∧Q→R) ↔ (P∨Q→R)
T53
P∨Q ↔ Q∨P
"commutative law for ∨"
T54
P ∨ (Q∨R) ↔ (P∨QR) ∨ R
"associative law for ∨"
T55
(P → Q∨R) ↔ (P→Q) ∨ (P→R)
"distribution of → over ∨"
T56
(P→Q) → (R∨P → R∨Q)
T57
(P→Q) → (P∨R → Q∨R)
T58
(P→Q) ∨ (Q→R)
Copyrighted material
"distribution of → over ∧"
"non-contradiction"
"separation of cases"
Chapter 2 -- 45
Version of Aug 2013
CHAPTER 2 SUMMARIES
T59
P ∨ ~P
T60
(P→R) ∨ (Q→R) ↔ (P∧Q → R)
T61
P ∧ (Q∨R) ↔ (P∧Q) ∨ (P∧R)
"distribution"
T62
P ∨ (Q∧R) ↔ (P∨Q) ∧(P∨R)
"distribution"
T63
P ∧ Q ↔ ~(~P∨~Q)
"de morgans"
T64
P ∨ Q ↔ ~(~P∧~Q)
"de morgans"
T65
~(P∧Q) ↔ ~P ∨ ~Q
"de morgans"
T66
~(P∨Q) ↔ ~P ∧ ~Q
"de morgans"
T67
~P∧~Q → ~(P∨Q)
T68
P ↔ (P∧Q) ∨ (P∧~Q)
T69
P ↔ (P∨Q) ∧(P∨~Q)
T70
Q → (P∧Q ↔ P)
T71
~Q → (P∨Q ↔ P)
T72
(P→Q) ↔ (P∧Q ↔ P)
T73
(P→Q) ↔ (P∨Q ↔ Q)
T74
(P↔Q)∧ P → Q
T75
(P↔Q)∧ Q → P
T76
(P↔Q)∧ ~P → ~Q
T77
(P↔Q)∧ ~Q → ~P
T78
(P → (Q↔R)) ↔ ((P→Q) ↔ (P→R))
T79
(P → (Q↔R)) ↔ (P∧Q ↔ P∧R)
T80
(P↔Q) ∨ (P↔~Q)
T81
(P↔Q) ↔ (P→Q)∧ (Q→P)
T82
(P↔Q) ↔ ~((P→Q) → ~(Q→P))
T83
(P↔Q) ↔ (P∧Q) ∨ (~P∧~Q)
T84
P∧Q → (P↔Q)
T86
((P↔Q) → R) ↔ (P∧Q → R)∧ (~P∧~Q → R)
T87
~(P↔Q) ↔ (P∧~Q)∨(~P∧~Q)
T88
P∧~Q → ~(P↔Q)
T89
~P∧Q → ~(P↔Q)
T90
~(P↔Q) ↔ (P↔~Q)
T91
P↔P
T92
(P↔Q) ↔ (Q↔P)
T93
(P↔Q) ∧(Q↔R) → (P↔R)
T94
(P ↔ (Q↔R)) ↔ ((P↔Q) ↔ R)
Copyrighted material
"excluded middle"
Chapter 2 -- 46
Version of Aug 2013
CHAPTER 2 SUMMARIES
T95
(P→Q) ↔ ((P↔R) ↔ (Q↔R))
T96
(P↔Q) ↔ (~P↔~Q)
T97
(P↔R)∧ (Q↔S) → ((P→Q) ↔ (R→S))
T98
(P↔R)∧ (Q↔S) → (P∧Q ↔ R∧S)
T99
(P↔R)∧ (Q↔S) → (P∨R ↔ Q∨S)
T100
(P↔R)∧ (Q↔S) → ((P↔Q) ↔ (R↔S))
T101
(Q↔S) → ((P→Q) ↔ (P→S)) ∧((Q→P) ↔ (S→P))
T102
(Q↔S) → (P∧Q ↔ P∧S)
T103
(Q↔S) → (P∨Q ↔ P∨S)
T104
(Q↔S) → ((P↔Q) ↔ (P↔S))
T105
P∧ (Q↔R) → (P∧Q ↔ R)
T106
(P → (Q→R)) ↔ ((P→Q) → (P→R))
T107
(P →→R)) ↔ (Q → (P→R))
T108
(P →(P→Q)) ↔(P→Q)
T109
((P→Q) → Q) ↔ ((Q→P) → P)
T110
P ↔ ~~P
T111
(P→Q) ↔ (~Q→~P)
T112
(P→~Q) ↔ (Q→~P)
T113
(~P→Q) ↔ (~Q→P)
T114
(~P→P) ↔ P
T115
(P→~P) ↔ ~P
T116
(P∧Q) ∨ (R∧S) ↔ (P∨R) ∧(P∨S) ∧(Q∨R) ∧(Q∨S)
T117
(P∨Q) ∧(R∨S) ↔ (P∧R) ∨ (P∧S) ∨ (Q∧R) ∨ (Q∧S)
T118
(P∨Q) ∧(R∨S) ↔ (~P∧~R) ∨ (~P∧S) ∨ (Q∧~R) ∨ (Q∧S)
T119
(P∨~P) ∧Q ↔ Q
T120
(P∧~P) ∨ Q ↔ Q
T121
P ∨ (~P∧Q) ↔ P∨Q
T122
P ∧ (~P∨Q) ↔ P∧Q
T123
P ↔ P ∨ (P∧Q)
T124
P ↔ P ∧ (P∨Q)
T125
(P→ Q∧R) → (P∧Q ↔ P∧R)
Copyrighted material
Chapter 2 -- 47
Version of Aug 2013
CHAPTER 2 Answers to the Exercises
Answers to the Exercises -- Chapter 2
SECTION 1
Exercises 1 and 2 answered together:
a.
Not a sentence
b.
Informal notation
~Q↔~R
F
/\
~Q ~R
F
T
|
|
Q
R
T
F
c.
Official notation
~(Q↔R)
T
|
Q↔R
F
/\
Q R
T F
d.
e.
Not a sentence
Informal notation
(P→Q) ∨ (R→~Q)
T
/\
P→Q
R→~Q
T
/\
/\
P Q
R ~Q
F
|
Q
f.
g.
Not a sentence
Informal notation
P∧Q → (Q→R∨Q)
T
/\
P∧Q Q→R∨Q
/\
T
P Q
/\
Q R∨Q
T
/\
R Q
T
Copyrighted material
Chapter 2 -- 48
Version of Aug 2013
CHAPTER 2 Answers to the Exercises
h.
Informal notation
P ↔ (P↔Q∧R)
F
/\
P P↔Q∧R
T
F
/\
P Q∧R
T
F
/\
Q R
F
i.
Informal notation
P ∨ (Q→P)
T
/\
P Q→P
T
/\
Q P
SECTION 2
1. a.
b.
c.
d.
e.
R∧P
W∨R
~R ∧ T
R∧S
Q↔R
2 and 3 answered together
a.
S∨V
false
b.
R↔S
true
c.
R∧S
false
d.
Q∨T
false
e.
Q∧S
false
SECTION 3
1. a.
~(P ∨ (Q ∧ R))
F
|
P ∨ (Q ∧ R)
T
/\
P Q∧R
T
/\
Q R
b.
~P ∨ (Q ∧ R)
F
/\
~P Q∧R
F
F
|
/\
Copyrighted material
Chapter 2 -- 49
Version of Aug 2013
CHAPTER 2 Answers to the Exercises
P Q R
T F F
c.
~(P ∨ R) ↔ ~P ∨ R
T
/\
~(P ∨ R) ~P ∨ R
F
F
|
/\
P∨R
~P R
T
F F
/\
|
P R
P
T
T
d.
~Q ∧ (P ∨ (Q↔R))
T
/\
~Q P ∨ (Q↔R)
T
T
|
/\
Q P Q↔R
F
T
/\
Q R
F F
e.
P → (~Q ↔ (~R → Q))
F
/\
P ~Q ↔ (~R → Q)
T
F
/\
~Q ~R → Q
T
F
|
/\
Q
~R Q
F
T F
|
R
F
2. a.
b.
c.
d.
3. a.
~V ↔ ~W; "won't" is a negation with narrow scope.
~V → (Y→W∧V); "both" gives rise to a conjunction with narrow scope since it splits the names
from the predicate. The comma prevents Y and V from occurring together.
Y ∨ (W ∨ V); "unless" is a disjunction sign.
(Y ∧ ~V) ∨ (V ∧ ~W)
Only if Veronica doesn't leave will William leave, or Veronica and William and Yolanda will all leave.
(Only if Veronica doesn't leave will William leave) ∨ (Veronica and William and Yolanda will leave)
(William will leave → Veronica doesn't leave) ∨ (V ∧ W ∧ Y)
(W → ~V) ∨ (V ∧ W ∧ Y)
b.
If neither William nor Veronica leaves, Yolanda won't either
If neither William [leaves] nor Veronica leaves, [then] Yolanda won't [leave]
~(W ∨ V) → ~Y
Copyrighted material
Chapter 2 -- 50
Version of Aug 2013
CHAPTER 2 Answers to the Exercises
c.
If William will leave if Veronica leaves, then he will surely leave if Yolanda leaves
If (William will leave if Veronica leaves) then ([William] will leave if Yolanda leaves)
(V → W) → (Y → W)
d.
Neither William nor Veronica nor Yolanda will leave
~(W ∨ V ∨ Y)
4.
" Veronica leaves but neither William nor Yolanda leaves" corresponds to the truth-value
assignment: V --- true; W --- false; Y --- false. We use parse trees to compute the truth values of the
complex sentences.
a.
(W → ~V) ∨ (V ∧ W ∧ Y)
T
/\
W → ~V V ∧ W ∧ Y
T
F
/\
/\
W ~V
V∧W Y
F F
F
F
|
/\
V
V W
T
T F
b.
~(W ∨ V) → ~Y
T
/\
~(W ∨ V) ~Y
F
T
|
|
W∨V
Y
T
F
/\
W V
F T
c.
(V → W) → (Y → W)
T
/\
V→W Y→W
F
T
/\
/\
V W Y W
T F
F F
d.
~(W ∨ V ∨ Y)
F
/\
W∨V∨Y
T
/\
W∨V Y
T
F
/\
W V
F
T
Copyrighted material
Chapter 2 -- 51
Version of Aug 2013
CHAPTER 2 Answers to the Exercises
5. a.
Sally will run and win unless she quits
(Sally will run and [Sally will] win) ∨ ([Sally] quits)
(R ∧ W) v Q
b.
Sally will win exactly in case she runs without quitting
Sally will win exactly in case (she runs [and doesn't] quit)
W ↔ (R ∧ ~Q)
c.
Sally, who will run, will win if she doesn't quit
Sally will run, and Sally will win if she doesn't quit
R ∧ (~Q → W)
d.
Sally will run and quit, but she will win anyway
Sally will run and quit, and she will win
(R ∧ Q) ∧ W
SECTION 4
1. a.
b.
c.
d.
e.
f.
g.
h.
i.
None; if we had ~~Q instead of Q it would be an instance of MTP.
Simplification
Double Negation
MTP
CB
None.
BC
Addition
None
2. a.
b.
c.
d.
e.
f.
~W ↔ ~X by CB; also ~X ↔ ~W by CB
~W by MTP
Nothing
~W by S; also ~X by S
W → ~X by BC; also ~X → W by BC
Nothing
SECTION 5 Derivations of numbered theorems not given
SECTION 6 Derivations of numbered theorems not given
SECTION 7
1. a.
b.
All fine
In line 8, the sentence that can be inferred from 7 by RT39 is W → ~S.
2, 3, 4, 5: Derivations of numbered theorems not given
SECTION 8
1. a.
b.
c.
All fine
Line 4: MTP does not apply;
Line 8: BC (biconditional to conditional) does not apply; we could use CB;
Line 11: MP does not apply to biconditionals; you have to split the biconditional into conditionals
first using BC.
Line 2: the result of applying DM to pr2 is ~Y ∧ ~~Z rather than ~Y ∧ Z.
Line 3: NC doesn't apply; the NC would generate line 3 if line 2 were Y ∧ ~Z.
Line 4: Line 4 is not available at line 4; it may not be cited to justify itself. The sentence could be
Copyrighted material
Chapter 2 -- 52
Version of Aug 2013
CHAPTER 2 Answers to the Exercises
generated by applying MT to line 3 and pr1.
2. a.
U∧V → X
~V → Y
X∨Y → Z
∴ ~Z → ~U
1. Show ~Z → ~U
2.
~Z
3.
~(X ∨ Y)
4.
~X ∧ ~Y
5.
~X
6.
~Y
7.
~(U ∧ V)
8.
~U ∨ ~V
9.
~~V
10.
~U
b.
1.
2.
3.
4.
5.
(X→Y) → Z
~Z
V→Y
∴ ~V
Show ~V
~(X→Y)
X ∧ ~Y
~Y
~V
c.
<use dm>
ass cd
pr3 2 mt
3 dm
4s
4s
pr1 5 mt
7 dm
6 pr2 mt
8 9 mtp cd
<use nc>
pr1 pr2 mt
2 nc
3s
4 pr3 mt dd
P∨Q
Q→S
U ∨ ~S
P∨S → R
R→U
∴ U
1. Show U
2.
Show P → U
3.
P
4.
P∨S
5.
R
6.
U
7.
Show Q → U
8.
Q
9.
S
10.
~~S
11.
U
12.
U
13.
Copyrighted material
ass cd
3 add
4 pr4 mp
5 pr5 mp cd
ass cd
8 pr2 mp
9 dn
10 pr3 mtp cd
pr1 2 7 sc
12 dd
Chapter 2 -- 53
Version of Aug 2013
CHAPTER 2 Answers to the Exercises
SECTION 9
1. a.
~(P ↔ Q)
R∨P
~Q → R
∴ R
1.
2.
3.
4.
5.
6.
7.
8.
Show R
~R
P
P ↔ ~Q
P → ~Q
~Q
R
b.
W→U
~W → V
∴ U∨V
Show U ∨ V
Show ~U → V
~U
~W
V
U∨V
1.
2.
3.
4.
5.
6.
c.
<use nb>
ass id
2 pr2 mtp
pr1 nb
4 bc
3 5 mp
6 pr3 mp
2 7 id
<use cdj>
ass cd
3 pr1 mt
4 pr2 mp cd
2 cdj dd
P ∨ (Q∧S)
R∨Q
S ∨ ~P
Q → ~S
∴ R
1. Show R
2.
~R
3.
Q
4.
~S
5.
~Q ∨ ~S
6.
~(Q∧S)
7.
P
8.
~~P
9.
S
10.
Copyrighted material
ass id
2 pr2 mtp
3 pr4 mp
4 add
5 dm
6 pr1 mtp
7 dn
8 pr3 mtp
4 9 id
Chapter 2 -- 54
Version of Aug 2013
CHAPTER 2 Answers to the Exercises
SECTION 10
(R↔S) ∨ (R↔~S); tautology
1. a.
(R↔S) ∨ (R↔~S)
T
T
T
T
R
T
T
F
F
S
b.
R ↔ (S↔R); not a tautology
R
T
S
F
c.
R ∨ (S∧T) → R ∧ (S∨T); not a tautology
R
F
S
T
d.
~U → (U→~V); tautology
U
T
T
F
F
V
T
F
T
F
e.
(~R↔R) → S; tautology
R
T
T
F
F
S
f.
(S∧T) ∨ (S∧~T) ∨ ~S; tautology
T
F
T
F
R ↔ (S↔R)
F
T
T
~U → (U→~V)
T
T
T
T
(~R↔R) → S
T
T
T
T
T
F
T
F
T
T
T
F
F
R ∨ (S∧T) → R ∧ (S∨T)
F
(S∧T) ∨ (S∧~T) ∨ ~S
T
T
T
T
S
T
F
T
F
SECTION 11
a.
U
T
U∧V → X
~V → U
X∨V → U
∴ V → ~U
V
T
X
T
NO
U∧V → X
T
Copyrighted material
~V → U
T
X∨V → U
T
V → ~U
F
Chapter 2 -- 55
Version of Aug 2013
CHAPTER 2 Answers to the Exercises
b.
X
T
T
T
T
F
F
F
F
Y
T
T
F
F
T
T
F
F
c.
P
T
T
T
T
F
F
F
F
S
T
U
T
T
T
T
F
F
F
F
R
T
F
T
F
T
F
T
F
~(P ↔ Q)
F
F
T
T
T
T
F
F
W
T
W
T
F
T
F
T
F
T
F
R
F
Copyrighted material
~Y
F
F
T
T
F
F
T
T
R∨P
T
T
T
T
T
F
T
F
~Q → R
T
T
T
F
T
T
T
F
R
T
F
T
F
T
F
T
F
~T ∨ ~S
T
~S
F
NO
S∨T
T
W∨S
T
YES
W→U
T
T
T
T
F
T
F
T
P ↔ ~Q
Q → R∨P
R → ~Q ∨ ~P
∴ Q∨R
Q
F
~Z
F
T
F
T
F
T
F
T
YES
W→U
~W → V
∴ U∨V
V
T
T
F
F
T
T
F
F
f.
(X→Y) → Z
T
F
T
T
T
F
T
F
S∨T
W∨S
~T ∨ ~S
∴ ~S
T
F
e.
Z
T
F
T
F
T
F
T
F
YES
~(P ↔ Q)
R∨P
~Q → R
∴ R
Q
T
T
F
F
T
T
F
F
d.
P
T
(X→Y) → Z
~Z
∴ ~Y
P ↔ ~Q
T
~W → V
T
T
T
F
T
T
T
F
U∨V
T
T
T
T
T
T
F
F
NO
Q → R∨P
T
R → ~Q ∨ ~P
T
Chapter 2 -- 56
Q∨R
F
Version of Aug 2013
CHAPTER 2 Answers to the Exercises
g.
P
T
T
T
T
F
F
F
F
Q
T
T
F
F
T
T
F
F
h.
P
T
P ∨ (Q∧S)
S∨Q
S ∨ ~P
∴ S
S
T
F
T
F
T
F
T
F
P ∨ (Q∧S)
T
T
T
T
T
F
F
F
P ∧ (Q∨S)
S∨Q
S∨P
∴ S
Q
T
S
F
P ∧ (Q∨S)
T
Copyrighted material
YES
S∨Q
T
T
T
F
T
T
T
F
S ∨ ~P
T
F
T
F
T
T
T
T
S
T
F
T
F
T
F
T
F
S∨P
T
S
F
NO
S∨Q
T
Chapter 2 -- 57
Version of Aug 2013
CHAPTER 3 SECTION 1
Chapter Three
Name letters, Predicates, Variables and Quantifiers
1 NAME LETTERS AND PREDICATES
In chapters 1 and 2 we studied logical relations that depend only on the sentential connectives: '~', '→', '∧',
'∨', '↔'. The atomic sentences -- those that contain no connectives -- were symbolized by sentential
letters, and we paid no attention to any internal structure that they might have. It is now time to study that
structure. The Predicate Calculus is a system of logic that studies the ways in which sentences are
constructed out of name letters, predicates, variables, and quantifiers, as well as connectives. We have
already studied connectives; in this section we introduce name letters, predicates, variables, and
quantifiers.
In our logical symbolism, name letters are written as the small letters: a, b, c, d, e, f, g, h (and with
subscripts, such as ‘c3’). Any small letter between 'a' and 'h' can be used as a name letter. Name letters
in the logical symbolism correspond to names of English:
Carlos, Agatha, Dr. Samuelson, Ms. Bernstein, Madame Curie, David Rockefeller, San
Diego, Germany, UCLA, General Electric, Microsoft, Google, Macy's, The Los Angeles
Times, I-405, Memorial Day, the FBI, ...
Any one of these may be symbolized by means of a name letter:
h
c
g
Henry
California
General Electric
The simplest way to make a sentence containing a name letter is to combine it with a one-place
predicate. One-place predicates appear in our logical symbolism as the capital letters from A to O (and
with subscripts, such as ‘G2’). One-place predicates correspond roughly to grammatical predicates in
English; in the following examples, the underlined phrases would be symbolized as one-place predicates:
Agatha is clever.
Henry is a giraffe.
Ferdy dances well.
Georgia is a state
Ann will run for re-election.
(The parts that are not underlined are symbolized with name letters.)
Whereas English proper names are usually capitalized, the logical name letters that represent them are
not, and whereas English predicates are typically not capitalized, the logical predicates that represent
them are capitalized. There is nothing "logical" about this reverse convention; it is an historical accident,
but it has now become part of the tradition of symbolic logic Further, in the usual formulations of the
predicate calculus the predicate comes before the name letter, instead of after it as in English. This, too,
is an historical accident. So the sentences given above can be symbolized as follows:
Agatha is clever.
Henry is a giraffe.
Ferdy dances well.
Georgia is a state.
Ann will run for re-election.
Ca
Gh
Df
Ag
Ea
A one-place ("monadic") predicate is any capital letter between 'A' and 'O' (optionally with a
numerical subscript).
A name letter is any small letter from 'a' to 'h' (optionally with a numerical subscript).
An atomic sentence may be formed by writing a one-place predicate followed by a name
letter.
Chapter Three -- 1
Version of Aug 2013
CHAPTER 3 SECTION 1
EXERCISES
1. Symbolize each of the following sentences:
a. Fred is an orangutan.
b. Gertrude is an orangutan but Fred isn't.
c. Tony Blair will speak first.
d. Gary lost weight recently; he is happy.
e. Felix cleaned and polished.
f. Darlene or Abe will bat clean-up.
We assume that a one-place predicate is true of certain things, and that a name letter stands for a unique
thing. A sentence consisting of a one-place predicate together with a name letter is true if and only if the
predicate is true of the thing that the name letter stands for. Thus, taking the examples listed above, we
assume that 'C' is true of all and only clever things, that 'a' stands for Agatha (presumably a person or
animal), and then:
Ca
is true if and only if Agatha is one of the clever things that the predicate is true of. Similarly, if `G' is true of
giraffes, then `Gh' is true if Henry is one of the giraffes. If `E' is true of the things that will run for
re-election, and if 'a' stands for Ann, then `Ea' is true if and only if Ann will run for reelection.
Predicates are generally true of several specific things, but a predicate might be true of only one thing ('is
a moon of the earth') or might not be true of anything at all. If there are in fact no dragons, the sentence:
Df
Fred is a dragon
contains a predicate 'D' that is true of nothing at all. This means that the sentence `Df' will be false, no
matter who or what `Fred' stands for.
In this chapter we assume that each name letter in our logical symbolism stands for a unique thing. This
assumption is an idealization, for it is not true that the words of English that we are representing by name
letters always succeed in naming something. If there is no such person as Paul Bunyan, then `Paul
Bunyan' is a "name" that names nothing at all. In some systems of logic it is possible to use name letters
which do not stand for anything; these systems of logic are called "free logics". (They are called "free"
because they are "free of" the assumption that the name letters they contain actually stand for things.)
Free logics are a bit more complicated than standard logic. (Studies of free logic assume that the reader
is already acquainted with the standard logic taught here.) In this text we assume that any name letter
that we use stands for something.
EXERCISES
2. Symbolize each of the following, assuming:
`D' is true of doctors
'L' is true of people who are in love
'h' stands for Hans
'a' stands for Amanda
a.
b.
c.
d.
f.
Hans is a doctor but Amanda isn't.
Hans, who is a doctor, is in love
Hans is in love but Amanda isn't
Neither Hans nor Amanda is in love
Hans and Amanda are both doctors.
Chapter Three -- 2
Version of Aug 2013
CHAPTER 3 SECTION 1
3. Symbolize each of the following, using:
'L' for things that live in Brea
'D' for things that drive to school
a.
b.
c.
d.
e.
Eileen and Cosi both live in Brea.
Eileen drives to school, and so does Hank.
If Hank lives in Brea then he drives to school; otherwise he doesn't drive to school.
If David and Hank both live in Brea then David drives to school but Hank doesn't.
Neither Hank nor Eileen live in Brea, yet each of them drives to school.
2 QUANTIFIERS, VARIABLES, AND FORMULAS
So far, we have no means at all in our symbolism to express generalities. We can say that Pedro is a
doctor, and we can say that Pedro is wealthy, but we cannot say that everyone is a doctor, or that every
doctor is wealthy. Nor can we deny that everyone is a doctor, or say that some doctor isn't wealthy. We
cannot even express these claims. In order to express generalities we will introduce quantifiers and
variables.
Variables: Any small letter from 'i' to 'z' is a variable; also small letters between
'i' and 'z' with numerical subscripts.
The universal quantifier sign is '∀'.
The existential quantifier sign is '∃'.
A quantifier is either quantifier sign followed by a variable:
∀x
∀z
∀s
∃x
∃z
∃s
Here is how we use quantifiers. Suppose that we wish to say -- as some philosophers have said -- that
everything in the universe is either mental or physical. Suppose that `M' is the one-place predicate `is
mental', and `H' is the one-place predicate `is physical'. Then we symbolize the claim that everything is
either mental or physical as follows:
∀x(Mx ∨ Hx).
The initial `∀x' is a universal quantifier phrase. This is followed by something, `(Mx ∨ Hx)', which we will
call a symbolic formula. A formula is just like a symbolic sentence except that instead of a name letter
following each predicate we may have a variable, such as `x' above. The displayed formula says that
everything satisfies a certain condition. The universal quantifier is responsible for the "everything" part,
and the combination of variables and predicates tells us what the condition is. In the case in point, the
condition is that it is either mental or physical:
∀x (Mx ∨ Hx)
Everything
is such that
it is either mental or physical
Chapter Three -- 3
Version of Aug 2013
CHAPTER 3 SECTION 2
An existential quantifier can appear in a formula in the same place that a universal quantifier may appear:
∃x (Mx ∨ Hx)
Something
is such that
it is either mental or physical
In order to construct sentences in our new extended notation, we begin by defining what a symbolic
formula is. Intuitively, a symbolic formula is like a sentence, except that it may contain variables in places
where name letters otherwise would appear. We use the word 'term' to cover both name letters and
variables.
Terms: Any name letter or variable is a term.
So 'a' and 'x' are both terms. A formula is built up in steps, as follows:
Sentence letters: Any sentence letter is an atomic formula.
Atomic formulas: A one-place predicate followed by a term is an atomic formula.
Thus, if F is a one-place predicate and 'a' is a name letter, then 'Fa' is an atomic
formula;
If 'F' is a one-place predicate and 'x' is a variable then 'Fx' is an atomic formula.
Both 'Henry is a giraffe' and 'x is a giraffe' are symbolized as atomic formulas:
Gh
Gx
Molecular formulas: If □ and ○ are formulas, then the following are
molecular formulas:
~□
(□∧○)
(□∨○)
(□→○)
(□↔○)
Here are some molecular formulas:
~Gh
~Gx
(Gx ∧ Fa)
(Gx ∨ Jc)
(Gh → Jy)
(~Fa ↔ Ga) → Hx
We can also make formulas out of other formulas by "generalizing" them with quantifiers:
Quantified formulas: If □ is a formula, and 'x' is a variable, then these are quantified
formulas:
∀x□
∃x□
Examples of quantified formulas are:
∀xGx
∃xFx
∀y(Gy→Fy)
∃w~(Gw ∧ ~Fb)
∀v(~Jx ↔ Fv)
Once a quantified formula is constructed, it may be used as input to any of these provisions. So, given
that the examples above are formulas, we can make new formulas by combining them with connectives:
(∀xGx ∧ ∃xFx)
(∃xFx ∨ ∀y(Gy→Fy))
∀y(Gy→Fy)
Chapter Three -- 4
(P ∧ ∃xFx)
Version of Aug 2013
CHAPTER 3 SECTION 2
We may informally omit parentheses exactly as we did in the last chapter, to produce informal notation:
∀xGx ∧ ∃xFx
∃xFx ∨ ∀y(Gy→Fy)
(Note that '∀yGy→Fy', is a conditional; it is not equivalent to '∀y(Gy→Fy)', which is a universal
generalization of a conditional.)
Likewise, we can add a quantifier to a formula that already has one or several quantifiers within it:
∀x(Gx → ∃yFy)
∀x~∃y(Gx ∨ ~Fy)
∀x∀y∀z(Gx → Fz)
A formula is anything that can be constructed by means of the above
provisions for atomic formulas, molecular formulas, and quantified formulas.
Nothing else is a formula.
Every formula is either atomic, or it has a main connective or a quantifier with scope over the whole
formula. The main connective or quantifier in a formula is the last connective or quantifier that was added
in constructing the formula. Formulas may be parsed as in chapters 1 or 2. Some examples are:
∀x(Gx → ∃yFy)
|
(Gx → ∃yFy)
2
Gx
∃yFy
|
Fy
∀x~∃y(Gx ∨ ~Fy)
|
~∃y(Gx ∨ ~Fy)
|
∃y(Gx ∨ ~Fy)
|
(Gx ∨ ~Fy)
2
Gx
~Fy
|
Fy
EXERCISES
1. For each of the following, say whether it is a formula in official notation, or in informal notation, or not a
formula at all. If it is a formula, parse it.
a. ~∀x(Fx → (Gx ∧ Hx))
b. ∃x~~Gx → Hx ∨ ∃yGy
c. ~(Gx ↔ ~Hx)
d. ∀xGx ∧ ∃Hx
e. Fa → (Gb ↔ Hc)
f. ∀x(Gx ↔ x ∨ Ha)
g. ∀x(Gx ↔ Hx) → Ha ∧ ∃zKz
Chapter Three -- 5
Version of Aug 2013
CHAPTER 3 SECTION 3
3 SCOPE AND BINDING
In the following we will need to distinguish a symbol from an occurrence of that symbol. For example, the
formula:
∀xFx
contains one variable, the variable 'x', which occurs twice in the formula. It has one occurrence as part of
the quantifier, and one occurrence immediately following the predicate 'F'. It will be important to be able to
say when an occurrence of a quantifier binds an occurrence of a variable. This can be given a precise
explanation in terms of the scope of an occurrence of a quantifier. The scope of an occurrence of a
quantifier includes itself along with the formula to which it was prefixed when constructing the whole
formula. Here are some occurrences of quantifiers and their scopes, indicated by underlining. (The line
immediately under a quantifier occurrence indicates its scope.)
∀x Fx
∀x(Fx → Gx)
∃xFx ∧ ∃y(Gy ∧ Hy)
∃x(Fx ∧ ∀yGy)
∃x(Fx ∧ ∃y(∃zGz ∧ Hy))
Using the notion of the scope of a quantifier, we can say when a quantifier occurrence binds an
occurrence of a variable in a formula:
A quantifier occurrence binds an occurrence of a variable if
the variable occurrence is within the scope of the quantifier occurrence
the variable occurrence is the same letter as the one in the quantifier occurrence
the variable occurrence is not already bound by another quantifier occurrence within
the scope of the first quantifier occurrence
(Notice that a variable occurrence that is part of a quantifier is automatically bound by that quantifier.)
The arrows here indicate which variables are bound by the quantifier:
∀x(Fx → Gx)
The initial quantifier binds both occurrences of 'x' because (1) they are within its scope, (2) they are the
same letter as the one in the quantifier itself, and (3) they are not already bound by another quantifier in
the formula. These examples are similar:
∃xFx ∧ ∃y(Gy ∧ Hy)
∃x(Fx ∧ ∀yGy)
∃x(Fx ∧ ∃y ( ∃zGz ∧ Hy))
Chapter Three -- 6
Version of Aug 2013
CHAPTER 3 SECTION 3
∃x(Fx ∧ ∃y ( ∃zGz ∧ Hy ∧ Hx))
The following example illustrates a case in which an occurrence of 'x' (the last one) is not bound by the
initial quantifier '∃x′, even though it is within its scope. This is because there is another quantifier inside
that already binds that occurrence of 'x':
∃x(Fx ∧ ∃x ( ∃zGz ∧ Hx))
Using the notion of a quantifier binding an occurrence of a variable, we can define what a sentence is:
A sentence is any formula in which every occurrence of a variable in the formula is
bound by an occurrence of a quantifier in the formula.
A variable occurrence that is not bound is called "free". So a sentence can also be defined as a formula
that contains no free occurrences of variables.
All of the examples given above are sentences. The following formulas are not sentences because
certain occurrences of variables in them are not bound any of their quantifiers:
∀x(Fy → Gx)
no quantifier contains 'y'
∃xFx ∧ ∃y(Gx ∧ Hy)
the scope of the initial quantifier does not include the second 'x'
∃x(Fx ∧ ∃y (∃zGz ∧ Hz))
the scope of the quantifier with 'z' does not extend far enough
∃x(Fx ∧ ∃x(∃zGz ∧ Hy))
no quantifier contains 'y'
EXERCISES
1. For each of the following, say whether it is a sentence, a formula that is not a sentence, or not a
formula at all. (Include sentences and formulas in informal notation as sentences and formulas.) If it is a
sentence or formula, indicate which quantifiers bind which variables.
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
∃x(Fx ∧ ∀y(Gy ∨ Hx))
∃y(Hy ∧ ∃~zHz)
∃z~(Hz ∧ Gx ∧ ∃xIx)
~(~Gx → ∀y(Jx ∧ Ky ↔ Lx))
∃xGx ↔ ∃y(Gy ∧ Hx)
∀x(Gx → ∀y(Hy → ∀z(Iz → Hx ∧ Gz)))
∀x∃y(Hx ↔ ~Gy)
∀xy(Gx ∧ Hy → Kx)
∀x(Gx ∧ ∃y → Hx ∧ Jy)
∀x∃y∀z(Gx ↔ ∃w(Hw ∧ ~Hx ∧ Gy))
Chapter Three -- 7
Version of Aug 2013
CHAPTER 3 SECTION 4
4 MEANINGS OF THE QUANTIFIERS
What do quantifiers mean? This can be answered indirectly by giving a way to read symbolic formulas in
English. We already know how to read the parts of formulas without quantifiers or variables; we have:
Gh
Ea
Gh ∧ Ea
Gh → Ea
Henry is a giraffe
Ann with run for reelection
Henry is a giraffe and Ann will run for reelection.
If Henry is a giraffe then Ann will run for reelection.
We can read a quantified formula by adding these provisions:
Read any universal quantifier as "everything is such that", while reading any variable that it
binds as a pronoun which has the 'everything' as its antecedent.
Read any existential quantifier as "something is such that" while reading any variable that it
binds as a pronoun which has the 'something' as its antecedent.
Here are some examples:
∀xGx
∃x(Gx ∧ Ex)
∀x(Gx → Ex)
everything is such that it is a giraffe
something is such that it is a giraffe and it will run for reelection
everything is such that if it is a giraffe then it will run for reelection
These readings are stilted, and sometimes cumbersome. But they are accurate paraphrases of the
symbolic notation. Often there are more natural ways to word an English sentence. For example, these
are all equivalent:
∃x(Gx ∧ Ex)
something is such that it is a giraffe and it will run for reelection
something is a giraffe which will run for reelection
some giraffe will run for reelection
Likewise, these are all equivalent:
∀x(Gx → Ex)
everything is such that if it is a giraffe then it will run for reelection
everything, if it is a giraffe, will run for reelection
every giraffe will run for reelection
As in the case of connectives, we need to distinguish carefully between the official definition of the
quantifiers and the question of how best to read them in English. The official definition of the quantifiers
has to do with the truth-values of the sentences that are produced using them:
Definitions of the quantifiers
To tell whether or not a sentence of the form ∀x(...x...x...) is true:
Remove the initial universal quantifier. Pretend that the variable it was binding is a
name letter. If you now have a sentence that is true no matter what the pretend
name stands for, then the original sentence is true; otherwise it is false.
To tell whether or not a sentence of the form ∃x(...x...x...) is true:
Remove the initial existential quantifier. Pretend that the variable it was binding is a
name letter. If there is something that the pretend name could stand for such that
the sentence you now have is true, then the original sentence is true; otherwise it is
false.
Chapter Three -- 8
Version of Aug 2013
CHAPTER 3 SECTION 4
To apply this to the example 'Everything is either mental or physical':
Begin with the sentence:
∀x(Mx ∨ Hx).
Erase the initial quantifier, yielding:
Mx ∨ Hx.
Now pretend that `x' is a name letter, and ask ourselves:
Is `Mx ∨ Hx' true no matter what `x' stands for?
If the answer is yes, then the original sentence `∀x(Mx ∨ Hx)' is true;
otherwise `∀x(Mx ∨ Hx)' is false.
This test explains why we read `∀x(Mx ∨ Hx)' in English as `Everything is either mental or physical'. It is
because the test for the truth of `∀x(Mx ∨ Hx)' succeeds if everything is indeed either mental or physical,
and it fails if not everything is either mental or physical. To see that this is so, compare the meaning of the
English sentence with the official statement of the conditions under which the symbolized version is true:
Suppose that certain philosophers are right, and everything is either mental or physical. Then if
we treat `x' as a name letter, the phrase `Mx ∨ Hx' must be true no matter what `x' stands for.
Because it can only stand for something that is mental or physical (that's all there is), and if it
stands for something mental the first disjunct is satisfied, and if it stands for something physical
then the second disjunct is satisfied.
Suppose on the other hand that not everything is either mental or physical. (Suppose, as some
philosophers have argued, that the number 4 is neither a mental thing nor a physical thing.) Then
if we treat `x' as a name letter, we will not find that the phrase `Mx ∨ Hx' is true no matter what `x'
stands for. For if `x' stands for the number 4, neither disjunct will be satisfied.
These considerations do not settle the question of whether everything is either mental or physical. Instead
they show that there is an equivalence between the truth-value, in English, of the sentence `Everything is
either mental or physical', and the truth-value, according to our official account, of the predicate calculus
sentence `∀x(Mx ∨ Hx)'.
EXERCISES
1. Suppose that `A' stands for `is a sofa', `B' stands for `is well-built' and `C' stands for `is comfortable'.
For each of the following sentences, produce an accurate but "cumbersome" reading in English as well as
a natural idiomatic reading if possible.
a. ∃x(Ax ∧ Bx)
b. ∀x(Ax → Bx)
c. ∀x(Ax ∨ Bx)
d. ∃x~Ax
e. ∀y~Ay
f. ∀z(Az ∧ Bz → Cz)
g. ∃xCx ∧ ∀yBy
h. ∃x(Cx → ∀yBy)
2. Assume that all giraffes are friendly, and that some giraffes are clever and some aren't. What are the
truth-values of these sentences?
a. ∀x(Gx → Fx)
b. ∀x(Gx → Cx)
c. ∃x(~Fx ∧ Gx)
d. ∃y(Fy ∧ Cy)
e. ∃z(Gz ∧ Cz)
f. ∀x(Gx → ~Gx)
Chapter Three -- 9
Version of Aug 2013
CHAPTER 3 SECTION 5
5 SYMBOLIZING SENTENCES WITH QUANTIFIERS
5A CATEGORICAL SENTENCES
The ancient Greek philosopher Aristotle is generally credited with the invention of formal logic. He devised
a fairly complete and accurate study of the logical relations among sentences of a certain special sort.
These are called "categorical" sentences, and they include any sentence which has one of the following
forms (with Aristotle's titles):
Universal affirmative:
Particular affirmative:
Universal negative:
Particular negative:
Every A is B
Some A is B
No A is B
Some A is not B
These categorical sentences are only a few of the forms that can be represented in modern predicate
logic, but they are simple and basic, and their treatment provides a nice introduction to the symbolism.
A universal affirmative sentence of the form:
Every A is B
is represented in the predicate calculus as:
∀x(Ax → Bx).
You can judge the adequacy of this for yourself by comparing the reading of the symbolic version with the
English form; that is, compare:
Everything is such that if it is an A then it is B
with:
Every A is B.
The question to ask for logical purposes is: Is there any possible situation in which these two sentences
differ in truth-value? If they agree in all logically possible situations, then the proposed symbolization is a
good one; otherwise not. Here is some reasoning that suggests the symbolization is a good one:
Suppose that in some possible situation every A is B. Then, in that situation everything
will be such that if it's an A then it is B. Suppose on the other hand that not every A is B.
Then there will be something that is an A but is not B. So it won't be true that everything
is such that if it's an A it is B.
Traditionally, the main reservation expressed about this symbolization concerns a possible situation in
which there are no A's at all. Suppose that a naturalist is uncertain about whether or not there are any
friendly elephants, but is willing to assert:
Every friendly elephant is an herbivore.
Suppose that there are in fact no friendly elephants. Then is what the naturalist said true or false? If we
accept the proposed symbolization above, we will represent the naturalist as having said something true.
Let us see why this is so. The proposed symbolization is:
∀x(x is a friendly elephant → x is an herbivore),
that is:
∀x(Fx ∧ Ex → Hx).
If there are no friendly elephants, this sentence will be true, because, treating `x' as a name letter, the
following is true no matter what `x' stands for:
Fx ∧ Ex → Hx.
It is true because no matter what `x' stands for, the antecedent is false (because there are no friendly
elephants).
Chapter Three -- 10
Version of Aug 2013
CHAPTER 3 SECTION 5
Is that a proper treatment of the English sentence that was asserted? The consensus on this matter
seems to be "sometimes yes, sometimes no." That is, sometimes when we say "Every A is B" we
presuppose or imply that there are some A's, and sometimes we are neutral on this. In this text we will
always take the weaker interpretation, supposing that "Every A is B" does not commit you to there being
any A's. It is true, not false, if there are no A's. This is just a convention (a widely adopted one) for our
convenience. (If you want a version of 'Every A is B' that does commit you to there being A's, you can
write instead: '∃xAx ∧ ∀x(Ax → Bx)'.)
The particular affirmative form -- "Some A is B" -- is easy to symbolize; it gets represented as:
∃x(Ax ∧ Bx),
that is, "Something is such that it is both A and B."
Plural forms of categorical sentences are symbolized just like the singular forms:
All A's are B
Some A's are B
Every A is B
Some A is B
∀x(Ax → Bx)
∃x(Ax ∧ Bx)
This might seem wrong if you think that the use of the plural in English commits you to the view that there
is more than one A which is B. (The symbolized version has no such commitment.) The answer seems
to be that we sometimes use the plural to convey the thought that there is more than one A, but
sometimes we are neutral about this. In this text we will adopt the weaker interpretation, which makes
"Some A's are B" true whenever there is at least one A that is B.
The universal negative form is:
No A is B
There are two equally natural ways to symbolize this. One way depends on noticing that "No A is B" is
equivalent to saying "Every A is not B," which can be symbolized as:
∀x(Ax → ~Bx).
The other way is to notice that "No A is B" is equivalent to denying that "At least one A is B," and
symbolizing the sentence as:
~∃x(Ax ∧ Bx).
Soon we will be able to prove that these two forms are logically equivalent.
There are two traps to beware of when symbolizing categorical sentences. They both involve trying to
make the symbolizations of "universal" and "particular" sentences look alike. Suppose that we want to
symbolize:
Some dogs are brown.
It will not be correct to symbolize this as:
∃x(Dx → Bx),
that is:
Something is such that if it's a dog then it's brown.
This would be wrong because in some possible situations the symbolized version would differ in
truth-value from the English version. Consider a possible situation which is just like the actual one except
that all dogs are black, white, or grey. The English sentence 'Some dogs are brown' would be false in that
situation. But the symbolized version would be true in that situation. It would be true for the totally
irrelevant reason that not everything is a dog!!! Remember the official account of the existential quantifier;
the sentence:
Chapter Three -- 11
Version of Aug 2013
CHAPTER 3 SECTION 5
∃x(Dx → Bx)
is true if there is something to let `x' stand for which makes this true:
Dx → Bx.
But that's easy; just let `x' stand for some thing that is not a dog -- and then we have a conditional whose
antecedent is false. And such a conditional is true. The symbolized version is automatically true if there is
anything that isn't a dog, whereas the English sentence is not automatically true in such a situation. So
the symbolization is not a good one to use for that English sentence.
The other trap is to try to symbolize:
Every A is B
as:
∀x(Ax ∧ Bx).
For example, you might try to symbolize:
Every dog is a mammal
as:
∀x(Dx ∧ Mx).
It is easy to see that this cannot be a correct symbolization, for the English sentence is true, whereas the
symbolized version is false. The symbolized version says:
Everything is such that it is a dog and it is a mammal,
that is:
Everything is both a dog and a mammal.
But you are not both a dog and a mammal, so the symbolic sentence is false. So the symbolic sentence
is not a correct way to represent the English sentence we are trying to symbolize, `All dogs are mammals',
since the English sentence is true. The right way to translate the English sentence is the way discussed
above:
∀x(Dx → Mx).
EXERCISES
1. Symbolize these sentences.
a. Every handsome elephant is friendly.
b. No handsome elephant is friendly.
c. Some elephants are not handsome.
d. Some handsome elephants are friendly.
e. Each friendly elephant is handsome.
f. A handsome elephant is not friendly.
g. No friendly elephant is handsome.
Chapter Three -- 12
Version of Aug 2013
CHAPTER 3 SECTION 5
5B COMPLEX CATEGORICAL FORMS
Many sentences are constructed out of categorical forms. An example is:
Every brown dog is happy and well-fed
To symbolize this sentence, notice that the sentence in fact is a universal affirmative sentence; it just
happens to have a complex antecedent and a complex consequent. So begin by using the pattern for
universal affirmatives:
∀x(x is a brown dog → x is happy and well-fed)
Then complete the symbolization by filling in the details in the antecedent and consequent:
∀x(Bx∧Dx → Hx∧Fx)
(The combination Adjective + Noun, such as 'brown dog', gets symbolized as a conjunction. For the
cases under consideration in this text, that is always the way to symbolize a combination consisting of an
adjective modifying a noun.)
This example is similar:
Some brown dog isn't either happy or lively.
Its overall form is that of a particular affirmative:
∃x(x is a brown dog ∧ x isn't either happy or lively)
Its symbolization is then got by filling in the details in the conjuncts:
∃x(Bx ∧ Dx ∧ ~(Hx ∨Lx))
Some other examples like this are:
No dog is happy unless every dog is well-fed
∀x(x is a dog → ~x is happy) unless ∀x(x is a dog → x is well-fed)
∀x(Dx → ~Hx) ∨ ∀x(Dx → Fx)
Each dog is happy unless it isn't well-fed
∀x(x is a dog → x is happy unless x is not well-fed)
∀x(Dx → Hx ∨ ~Fx)
As we have seen, categorical sentences can themselves be combined with connectives. Another
example is:
If every dog is well-fed, and every dog is an animal, and every animal is happy, then every dog is
both well-fed and happy.
This is a complex of categorical sentences:
If ∀x(Dx → Fx) and ∀y(Dy → Ay) and ∀z(Az → Hz) then ∀z(Dz → Fz ∧ Hz)
that is:
∀x(Dx → Fx) ∧∀y(Dy → Ay) ∧∀z(Az → Hz) → ∀z(Dz → Fz ∧ Hz)
Sometimes a sentence is apparently ambiguous, but variable binding resolves the ambiguity. This
happens in the example
Each dog is happy unless it isn't well-fed
We decided above to include the 'unless' as part of the consequent of the quantified conditional. We
might try instead to make 'unless' be the major connective:
Chapter Three -- 13
Version of Aug 2013
CHAPTER 3 SECTION 5
∀x(x is a dog → x is happy) unless x isn't well-fed
∀x(Dx → Hx) ∨ ~Fx
However, this leaves the 'x' unbound by the quantifier. You have a formula that is not a sentence, and
there is no way to interpret the unbound occurrence of 'x'. Whenever a symbolization of an ordinary
meaningful English sentence ends up with a variable that is not bound by any quantifier, the symbolization
will not be correct.
EXERCISES
2. Suppose that `A' stands for `is a U.S. state', `C' for `is a city', `L' for `is a capital', and `E' for `is in the
Eastern time zone'. What are the truth values of these sentences?
a. ∀x(Cx → Lx)
b. ∃x(Cx ∧ Lx)
c. ∃x(Cx ∧ Lx ↔ Ex)
d. ∀x(Cx ∧ Ex → Ax)
e. ~∃x(Ax ∧ Ex)
f.
∃x(Cx ∧ Ex) ∧ ∃x(Cx ∧ ~Ex)
g. ∃x(Cx ∧ Ex ∧ Ax)
h. ~∃x(Cx ∧ ~Cx)
3. Symbolize the following sentences:
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.
m.
n.
o.
All giraffes are spotted.
All clever giraffes are spotted.
No clever giraffes are spotted.
Every giraffe is either spotted or drab.
Some giraffes are clever.
Some spotted giraffes are clever.
Some giraffes are clever and some aren't.
Some spotted giraffes aren't clever.
No spotted giraffe is clever but every unspotted one is.
Every clever spotted giraffe is either wise or foolhardy.
Either all spotted giraffes are clever, or all clever giraffes are spotted.
Every clever giraffe is foolhardy.
If some giraffes are wise then not all giraffes are foolhardy.
All giraffes are spotted if and only if no giraffes aren't spotted.
Nothing is both wise and foolhardy.
Chapter Three -- 14
Version of Aug 2013
CHAPTER 3 SECTION 5
5C "ONLY"
In chapter 1 we looked at how 'only' affects the symbolization of conditionals. The same word occurs in
connection with quantification. Consider the sentence:
Only dogs are happy
Reflection on what this says indicates that it could be symbolized the same as:
Any non-dog isn't happy
and thus as:
∀x(~Dx → ~Hx)
But intuitively the sentence is also equivalent to:
Anything that's happy is a dog
∀x(Hx → Dx)
Fortunately, we will be able to prove later that these two forms are equivalent.
Recall that the effect of 'only' on 'if' is to reverse antecedent and consequent. Something like that occurs
here too; compare the sentences:
All dogs are happy
Only dogs are happy
∀x(Dx → Hx)
∀x(Hx → Dx)
They look pretty much the same except that the antecedent and consequent of the quantified conditional
are switched.
Here are some examples of symbolizations of sentences using 'only':
Dogs can run, but only birds can fly.
∀x(Dx → Cx) ∧∀x(Fx → Bx)
Only birds can fly, but not all of them can.
∀x(Fx → Bx) ∧~∀x(Bx → Fx)
Dogs are happy and frisky; giraffes are happy, but only the well-fed ones are frisky.
∀x(Dx → Hx ∧ Fx) ∧∀x(Gx → Hx) ∧∀x(Gx ∧ Fx → Ex)
(Using 'E' for 'is well-fed'.)
Notice that the last conjunct is not symbolized as:
∀x(Fx → Gx ∧ Ex)
This would say that everything that is frisky is a well-fed giraffe, which is not what is intended. The point is
that among giraffes only the well-fed ones are frisky. The last conjunct could also be symbolized as:
∀x(Gx → (Fx → Ex))
The word 'only' can create ambiguity. Consider the sentence:
Only brown dogs are happy
This could be read as saying that everything that is happy is a brown dog:
∀x(Hx → Bx ∧ Dx)
or it could be read as saying that among dogs, every happy one is brown:
Chapter Three -- 15
Version of Aug 2013
CHAPTER 3 SECTION 5
∀x(Dx → (Hx → Bx))
Usually we don't notice such ambiguity since it is usually clear from context which is meant. Emphasis
also helps; saying "Only brown dogs are happy" indicates that among dogs, only the brown ones are
happy. Out of context, the sentence is simply ambiguous.
EXERCISES
4. Symbolize these sentences. If a sentence is ambiguous, give all pertinent symbolizations.
a. Only friendly elephants are handsome
b. If only elephants are friendly, no giraffes are friendly
c. Only the brave are fair.
d. If only elephants are friendly then every elephant is friendly
e. All and only elephants are friendly.
f. If every elephant is friendly, only friendly animals are elephants
g. If any elephants are friendly, all and only giraffes are nasty
h. Among spotted animals, only giraffes are handsome.
i. Among spotted animals, all and only giraffes are handsome
j. Only giraffes frolic if annoyed.
5D RELATIVE CLAUSES
Relative clauses modify nouns, as adjectives do, although relative clauses are typically more complex.
There are two sorts of relative clause: restrictive and non-restrictive, illustrated by:
Non-restrictive
Restrictive
Dogs, which are frisky, are cute
Dogs which are frisky are cute
Non-restrictive relative clauses do not affect the noun they follow; instead they are used to insert a
comment in addition to what the main sentence says. The main sentence of the non-restrictive example is
that dogs are cute, and the additional comment is that they are frisky. The entire sentence is used to
make both of these claims. If we want to capture the whole content of a sentence with a non-restrictive
relative clause the best we can do is to conjoin the two claims:
Dogs are frisky ∧ Dogs are cute
∀x(Dx → Fx) ∧∀x(Dx → Cx)
A restrictive relative clause restricts the content of the noun to which it is adjoined. In the restrictive
example above, it is frisky dogs that are said to be cute, not dogs in general. The symbolization is:
Dogs which are frisky are cute
∀x(Dx ∧ Fx → Cx)
You can usually tell a non-restrictive relative clause, for it is set off from its surroundings by commas
before and after it. When there are no commas, we assume in this text that the reading is restrictive.
Restrictive relative clauses are like adjectives, in that in logical form they are conjoined with the noun that
they modify. In the above example 'dogs which are frisky' becomes the conjunction 'Dx ∧ Fx'. When the
relative clause is more complex, it gives you something complex to conjoin to the part originating with the
noun that is modified. This is seen in:
Every dog which is neither cute nor frisky is not happy.
∀x(Dx ∧ ~(Cx ∨ Fx) → ~Hx)
Chapter Three -- 16
Version of Aug 2013
CHAPTER 3 SECTION 5
EXERCISES
5. Symbolize these sentences.
a. Every giraffe which frolics is happy
b. Only giraffes which frolic are happy
c. Only giraffes are animals which are long-necked.
d. If only giraffes frolic, every animal which is not a giraffe doesn't frolic.
e. Some giraffe which frolics is long-necked or happy.
f. No giraffe which is not happy frolics and is long-necked.
g. Some giraffe is not both long-necked and happy.
5E IMPLICIT UNIVERSAL QUANTIFIERS
In the symbolizations we have considered so far, symbolic universal quantifiers have originated naturally
from "universal" quantifier words of English. For example, the universal quantifier is often used in
symbolizing a sentence with one of the words 'each', 'every', 'all' in it, and the position of the English
quantifier word often corresponds to the position of the symbolic quantifier. In 'Every A is B' the English
sentence begins with 'every' and its symbolization begins with '∀x'.
Sometimes a universal quantification originates with an English indefinite article 'a' or 'an'. This happens
in:
A dog that is well-fed is happy.
This sentence is most naturally treated as conveying a universal claim, that any dog that is well-fed is
happy:
∀x(Dx ∧ Fx → Hx)
This is in spite of the fact that the indefinite article often conveys an existential claim, as in:
A girl left early
∃x(Gx ∧ Lx)
A good test for this is whether the indefinite article can be paraphrased by 'each'; this is natural in the first
example, but not in the second.
A more interesting case is when an indefinite article occurs inside a sentence, indicating a universal
quantification with scope over the whole sentence. This happens in:
If a dog is well-fed, it is happy
This appears to be a conditional of the form:
a dog is well-fed → it is happy
But that won't do, since there is nothing to bind the variable that comes from the 'it' in the consequent.
Instead, the indefinite article indicates a universal quantification of dog, with the rest of the sentence within
its scope. That is, it has the form:
∀x(x is a dog → (x is well-fed → x is happy))
∀x(Dx →( Fx → Hx))
This happens in the following two examples as well. In the first:
A giraffe is wise if and only if it's not foolhardy.
Chapter Three -- 17
Version of Aug 2013
CHAPTER 3 SECTION 5
This has a logical form something like:
Every giraffe is such that it is wise if and only if it is not foolhardy
∀x(Gx → (Wx ↔ ~Fx))
This sentence is similar:
A brown dog is frisky only if it is happy
Every brown dog is such that it is frisky only if it is happy
∀x(Bx ∧ Dx → (Fx → Hx))
The idea that indefinite phrases sometimes correspond to universal quantifiers with wide scope applies
also to plural indefinites -- to plural nouns or noun phrases which have no article or quantifier word before
them. An example is:
If dogs are well-fed, then they are happy
∀x(x is a dog → (x is well-fed → x is happy))
∀x(Dx → (Ex → Hx))
EXERCISES
6. Symbolize the following sentences.
a. If a giraffe is happy then it frolics unless it is lame.
b. A monkey frolics unless it is not happy.
c. Among giraffes, only happy ones frolic.
d. All and only giraffes are happy if they are not lame.
e. A giraffe frolics only if it is happy.
f. Only giraffes frolic if happy.
g. All monkeys are happy if some giraffe is.
h. Cute monkeys frolic.
i. Giraffes run and frolic if and only if they are blissful and exultant.
j. If those who are healthy are not lame, then if they are exultant, they will frolic.
k. Only giraffes and monkeys are blissful and exultant.
l. The brave are happy.
m. If a giraffe frolics, then no monkey is blissful unless it is.
n. Giraffes and monkeys frolic if happy.
Chapter Three -- 18
Version of Aug 2013
CHAPTER 3 SECTION 6
6 DERIVATIONS WITH QUANTIFIERS
Our first step in including quantificational sentences in derivations is to extend all of the rules from
chapters 1 and 2 to include formulas which have free variables. Although we continue to use derivations
for arguments consisting entirely of sentences, it will be essential to also allow formulas inside of the
derivations.
In this section we introduce three rules for quantifiers.
Rule ui (universal instantiation): The first rule is simple; it says that if everything satisfies a certain
condition, any particular thing satisfies that condition. That is, from any universally quantified formula one
may infer the result of removing the initial quantifier, and replacing every occurrence of the variable that it
was binding by a name letter or by a variable:
Rule ui: (universal instantiation):
∀x ...x...x...
∴ …b…b…
∀x ...x...x...
∴ …y…y…
Every occurrence of 'x' that '∀x' was binding must be replaced with the same name or
variable.
An example of this rule is to validate the argument from 'everything is either mental or physical' to
'Disneyland is either mental or physical':
∀x(Mx ∨ Px)
∴ Ma ∨ Pa
by rule ui
A more typical application would be to use rule ui to validate an inference like this:
Every giraffe is happy
Fido is a giraffe
∴ Fido is happy
∀x(Gx → Hx)
Gf
∴ Hf
A derivation using rule ui to validate this argument could go like this:
1. Show Hf
2. Gf → Hf
3. Hf
pr1 ui
pr2 2 mp dd
The universal instantiation step takes us from "everything is such that if it is a giraffe then it is happy" to "if
Fido is a giraffe then Fido is happy". Modus ponens does the rest.
In using rule ui the quantifier must be on the front of the formula and it must have scope over the whole
formula. If it has a narrower scope, then it is fallacious to apply the rule. For example, this inference is
not permitted:
∀xFx → Fg
∴ Fb → Fg
If everything is happy, Gertrude is happy
If Betty is happy, Gertrude is happy
Chapter Three -- 19
(logically true)
(not logically true)
Version of Aug 2013
CHAPTER 3 SECTION 6
Rule eg (existential generalization): The second rule is the reverse of the first, using the existential
quantifier instead of the universal. It says that if a particular thing satisfies a certain condition, then
something satisfies it. That is, from any formula one may infer the result of replacing some occurrences
of a name letter or a variable in it by a new variable, putting an existential quantifier on the front using that
variable.
Rule eg (existential generalization):
...b...b...
∴ ∃x…x…b…
...y...y...
∴ ∃x…x…y…
(You need not replace every occurrence of 'b' or of 'y' by 'x'.)
For example, if Fido is a brown dog, then something is a brown dog:
Bf ∧ Df
∴ ∃x(Bx ∧ Dx)
The existential quantifier that is put on the front must have scope over the whole formula. If the formula
you start with is in informal notation, you may need to restore the dropped parentheses before applying the
rule, as we did here.
Here is a little derivation that uses both of these rules. It validates the argument:
Every dog is happy
Fido is a dog
∴ Something is happy
∀x(Dx → Hx)
Df
∴ ∃xHx
1. Show ∃xHx
2.
3.
4.
Df → Hf
Hf
∃xHx
pr1 ui
2 pr2 mp
3 eg dd
There is a difference between Rules ui and eg. When using rule ui, you must replace every occurrence of
the variable that the initial quantifier binds with a name or variable. For example, you cannot do this:
∀x(Dx → Hx)
∴ Dx → Hb
That is:
Everything is such that if it is a dog then it is happy.
∴ If it is a dog then Bob is happy
Rule eg is different. When using rule eg you needn't replace all of the occurrences. For example, from:
Bob is happy or Bob is sad
you may infer
Something is such that Bob is happy or it is sad.
This conclusion looks odd, but it should be clear that it follows logically.
Chapter Three -- 20
Version of Aug 2013
CHAPTER 3 SECTION 6
Here is another example of a derivation using both of our new rules:
Fido is a dog
Every dog is happy
∴ Some dog is happy
1. Show ∃x(Dx ∧ Hx)
2.
3.
4.
5.
Df → Hf
Hf
Df ∧ Hf
∃x(Dx ∧ Hx)
pr2 ui
2 pr1 mp
pr1 3 adj
4 eg dd
There is a constraint on both of these rules: there must be no "capturing". If a new variable appears in the
conclusion of either rule that was not there previously, it must not be "captured" by a quantifier in the
formula. Specifically, if a new variable appears, none of its new occurrences may be bound by a quantifier
already in the formula. For example, this use of rule eg is not permitted:
Df ∧ ∀x(Hf → Gx)
∴ ∃x(Dx ∧ ∀x(Hx → Gx))  the universal quantifier captures the variable 'x' that replaces the
second 'f'
No capturing:
When using rule ui or rule eg a new variable must not be introduced if some of its
new occurrences are bound by a quantifier in the original formula.
You will not often encounter cases of capturing; they usually happen by accident. The possibility of
capturing can be avoided by always choosing a variable that does not already occur in the formula.
EXERCISES
1. Symbolize these arguments and produce derivations for them.
a.
The sky is blue
Everything that is blue is pretty
∴ Something is pretty
b.
Every hyena is grey.
Every hyena is an animal
Jenny is a hyena
∴ Some animal is grey
c.
If some hyena is grey, every hyena is grey
Every scavenger is grey
Jenny is a hyena and a scavenger
Kathy is a hyena
∴ Kathy is grey
Chapter Three -- 21
Version of Aug 2013
CHAPTER 3 SECTION 6
Rule ei (existential instantiation): Our third rule is rule ei (existential instantiation). It works just like
universal instantiation, except that (1) it applies to an existential quantifier, (2) you must instantiate to a
variable, not to a name letter, and (3) you must use a variable that has not already occurred in the
derivation or in any of the premises that have been cited in the derivation. This rule is meant to capture
the following kind of reasoning. Suppose that you are given the information:
Every dog is happy
Something is a dog
and you wish to infer that something is happy. You are not told that any particular named thing is a dog;
you just know that there are some. You might reason as follows:
By the second premise, there are some dogs. Call one of them "z". Then z is happy (by the first
premise), so something is happy.
What you did was to choose a label, 'z', for some dog, without specifying which dog it is. Then you made
inferences using that label, ending up with a conclusion that does not contain the label. The label was just
a device to reason with.
It was important that you chose a label that was not already assigned to something. If you used an
already existing name for the label, that could lead to fallacies. For example, consider this bad argument:
Every dog is happy
Something is a dog
Fluffy is a cat
∴ Some cat is happy
It would be wrong to reason like this:
By the second premise, there are some dogs. Call one of them "Fluffy". Then Fluffy is happy (by
the first premise). Also, Fluffy is a cat (third premise). So some cat is happy.
By using the name 'Fluffy' for one of the dogs you were implicitly assuming that Fluffy was a dog. That
assumption is not justified. Formally we get around such an unjustified assumption by using only variables
for labels, and by requiring that these variables are not already used for something else. We accomplish
this by requiring that the new variable not have occurred already in the derivation:
Rule ei: (existential instantiation):
∴
∃x ...x...x...
…y…y…
You must replace every occurrence of 'x' that '∃x' was binding.
The variable 'y' must not occur in the existentially quantified formula itself, or in
previous lines in the derivation, or in a premise that has been cited on a previous line.
Here now is a derivation using all of our new rules:
∀x(Bx ∧ Dx → Ex)
∃x(Dx ∧ Fx)
∀y(Fy → By)
∴ ∃z(Dz ∧ Ez)
Every brown dog is well-fed.
Some dog is frisky
Everything frisky is brown
∴ Some dog is well-fed
Chapter Three -- 22
Version of Aug 2013
CHAPTER 3 SECTION 6
1. Show ∃z(Dz ∧ Ez)
2.
3.
4.
5.
6.
7.
8.
9.
Du ∧ Fu
Fu → Bu
Bu
Bu ∧ Du
Bu ∧ Du → Eu
Eu
Du ∧ Eu
∃z(Dz ∧ Ez)
pr2 ei
pr3 ui
2 s 3 mp
2 s 4 adj
pr1 ui
5 6 mp
2 s 7 adj
8 eg dd
('u' has not already occurred in the derivation)
The reader should check to see that each of the new rules is properly used.
This derivation illustrates an important strategy rule. Often you will have an opportunity to apply ei to
introduce a variable, and then use ui to instantiate to that variable. In the derivation just given, ei
introduces 'u' on line 2 and ui is used twice to instantiate to 'u', on lines 3 and 6. The strategy rule is that
when this is a possibility, you should always apply rule ei before you apply rule ui.
Strategy hint: When using both ei and ui to instantiate to the same variable, apply
rule ei before rule ui.
This is because if you try using ui first, you will not then be able to use ei to instantiate to the same
variable, because the variable will not then be new. For example, suppose that you started the above
derivation with:
1. Show ∃z(Dz ∧ Ez)
2. Fu → Bu
3. Du ∧ Fu
pr3 ui
pr2 ei
Line 3 is fallacious because you have instantiated to 'u', but 'u' has already occurred in the derivation,
which violates the constraint that the variable used in ei must be new.
Here is a straightforward illustration of our three rules:
Every crook who steals a lot and doesn't get caught is affluent..
No crook who gets caught is affluent.
Some lucky crooks steal a lot.
Some crooks who aren't lucky don't steal a lot.
Every crook who isn't lucky gets caught.
Every crook who is lucky doesn't get caught.
∴ Some crooks are affluent and some aren't.
∀x(Cx ∧ Ex ∧ ~Gx → Ax)
∀x(Cx ∧ Gx → ~Ax)
∃x(Lx ∧ Cx ∧ Ex)
∃x(Cx ∧ ~Lx ∧ ~Ex)
∀x(Cx ∧ ~Lx → Gx)
∀x(Cx ∧ Lx → ~Gx)
∴ ∃x(Cx ∧ Ax) ∧∃x(Cx ∧ ~Ax)
(In doing this derivation recall that 'P ∧ Q ∧ R' is informal notation for '((P ∧ Q) ∧R)'.)
Chapter Three -- 23
Version of Aug 2013
CHAPTER 3 SECTION 6
1. Show ∃x(Cx ∧ Ax) ∧ ∃x(Cx ∧ ~Ax)
Lz ∧ Cz ∧ Sz
Cz ∧ Lz → ~Gz
Sz
Lz
Cz
~Gz
Cz ∧ Sz ∧ ~Gz → Az
Az
Cz ∧ Az
∃x(Cx ∧ Ax)
Cu ∧ ~Lu ∧ ~Eu
Cu ∧ ~Lu
Cu ∧ ~Lu → Gu
Gu
Cu ∧ Gu → ~Au
~Au
Cu ∧ ~Au
∃x(Cx ∧ ~Ax)
∃x(Cx ∧ Ax) ∧∃x(Cx ∧ ~Ax)
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
pr3 ei
pr6 ui
2s
2ss
2ss
5 6 adj 3 mp
pr1 ui
6 4 adj 7 adj 8 mp
6 9 adj
10 eg
pr4 ei
12 s
pr5 ui
13 14 mp
pr2 ui
13 s 15 adj 16 mp
13 s 17 adj
18 eg
11 19 adj dd
Notice that the ei step in line 2 precedes the ui steps in lines 3 and 8, and that the ei step in line 12
precedes the ui steps in lines 14 and 16.
EXERCISES
2. Here is a fallacious derivation to validate this argument:
∃x(Nx ∧ Ex)
∃x(Nx ∧ Ox)
∴ ∃x(Nx ∧ Ox ∧ Ex)
some number is even
some number is odd
some number is both odd and even
Identify the error in the derivation.
1. Show ∃x(Nx ∧ Ox ∧ Ex)
2.
3.
4.
5.
Nz ∧ Ez
Nz ∧ Oz
Nz ∧ Oz ∧ Ez
∃x(Nx ∧ Ox ∧ Ex)
pr1 ei
pr2 ei
2 s 3 adj
dd
3. Produce derivations for each of the following (be careful to obey the strategy rule just given):
a. theorem T202: ∴ ∀x(Fx → Gx) → (∃xFx → ∃xGx)
b. half of T203:
∴ ∃x~Fx → ~∀xFx
c. half of T204:
∴ ∀x~Fx → ~∃xFx
T201
T202
T203
T204
∀x(Fx → Gx) → (∀xFx → ∀xGx)
∀x(Fx → Gx) → (∃xFx → ∃xGx)
~∀xFx ↔ ∃x~Fx
~∃xFx ↔ ∀x~Fx
Chapter Three -- 24
Version of Aug 2013
CHAPTER 3 SECTION 7
7 UNIVERSAL DERIVATIONS
We have two instantiation rules, one for each quantifier, and we have a generalization rule for the
existential quantifier. It is customary and useful to have some kind of universal generalization rule as well.
For example, someone might want to reason as follows:
Every dog is a mammal
Every mammal is an animal
∴ Every dog is an animal
A natural approach might be like this. Let x be anything whatsoever. Instantiating the first premise tells us
that if x is a dog, it is a mammal; and instantiating the second premise tells us that if x is a mammal, it is
an animal. So using techniques from chapter 1, we may infer that if x is a dog, x is an animal. Now since
'x' was chosen to represent anything whatever, we can infer that everything is such that if it is a dog it is an
animal. That is, every dog is an animal.
What we want to capture is the idea that if you can show something for any arbitrarily chosen thing, it
holds for everything. Something like:
Dx → Ax
∴ ∀x(Dx → Ax)
because x can be anything at all
For technical reasons, this principle will be formulated not as a rule, but as a special kind of derivation. It
will take the form that if you want to show a universal claim, and you succeed in showing that it holds for a
variable, x, then if x is completely arbitrary, you may box and cancel the show line for the universal claim.
So the above reasoning will take this form: If you have a derivation of this form:
Show ∀x(Dx → Ax)
:::::
:::::
Dx → Ax
where x is completely arbitrary
Then you can box and cancel
Show ∀x(Dx → Ax)
:::::
:::::
Dx → Ax
ud
where x is completely arbitrary
The requirement that x be completely arbitrary is realized by the technical requirement that 'x' shall not
have occurred free anywhere in the derivation available from the show line, or in a premise cited in such a
line.
Chapter Three -- 25
Version of Aug 2013
CHAPTER 3 SECTION 7
The "ud" notation is the name of our new form of derivation:
Universal Derivation (UD):
If you have a derivation of the following form:
Show ∀x . . . x . . . x . . .
:::::
:::::
...x...x...
Then if there are no uncancelled show lines in between the first and last lines
displayed, and if 'x' does not occur free on any line in the derivation available
from the show line, or in any premise cited in such a line, you may box and
cancel, using the notation 'ud'.
The reasoning suggested above may now be incorporated into a derivation like this:
∀x(Dx → Mx)
∀y(My → Ay)
∴ ∀z(Dz → Az)
Every dog is a mammal
Every mammal is an animal
∴ Every dog is an animal
1. Show ∀z(Dz → Az)
2.
3.
4.
5.
6.
7.
Dz → Mz
Mz → Az
Show Dz → Az
Dz
Mz
Az
8.
pr1 ui
pr2 ui
ass cd
2 5 mp
3 6 mp
cd
4 ud
The reader should check that this derivation meets the conditions necessary for a ud derivation.
In a previous exercise we proved half of theorem 203. The other half of T203 is more difficult, but we can
do it using a universal derivation.
∴ ~∀xFx → ∃x~Fx
It is easy to begin the derivation, setting up a conditional derivation:
1. Show ~∀xFx → ∃x~Fx
2.
3.
~∀xFx
?????
ass cd
With no other guide, our strategy rules say to try id:
1. Show ~∀xFx → ∃x~Fx
2.
3.
~∀xFx
Show ∃x~Fx
ass cd
4.
~∃x~Fx ass id
5. ???
Again there is no clear way to proceed. Since we are trying to derive any contradiction, it is natural to try
to derive the unnegation of line 2:
Chapter Three -- 26
Version of Aug 2013
CHAPTER 3 SECTION 7
1. Show ~∀xFx → ∃x~Fx
2.
3.
~∀xFx
Show ∃x~Fx
4.
5.
~∃x~Fx
Show ∀xFx
6.
?????
ass cd
ass id
We can in fact show this universally quantified formula, by means of a universal derivation. We only need
to show the part folowing the quantifier: Fx.
1. Show ~∀xFx → ∃x~Fx
2. ~∀xFx
3.
Show ∃x~Fx
4.
5.
~∃x~Fx
Show ∀xFx
6.
7.
Show Fx
???
ass cd
ass id
The rest is easy by means of another indirect derivation:
1. Show ~∀xFx → ∃x~Fx
2.
3.
~∀xFx
Show ∃x~Fx
ass cd
4.
5.
~∃x~Fx
Show ∀xFx
ass id
6.
7.
8.
9.
Show Fx
~Fx
∃x~Fx
~∃x~Fx
ass id
7 eg
4r
We can now complete the universal derivation:
1. Show ~∀xFx → ∃x~Fx
2. ~∀xFx
3.
Show ∃x~Fx
ass cd
4.
5.
~∃x~Fx
Show ∀xFx
ass id
6.
Show Fx
7.
8.
9.
~Fx
∃x~Fx
~∃x~Fx
10.
11.
12.
ass id
7 eg
4 r 8 id
6 ud
~∀xFx
2 r 6 id
3 cd
Chapter Three -- 27
Version of Aug 2013
CHAPTER 3 SECTION 7
The rest is straightforward:
1. Show ~∀xFx → ∃x~Fx
2. ~∀xFx
3.
Show ∃x~Fx
ass cd
4.
5.
~∃x~Fx
Show ∀xFx
ass id
6.
Show Fx
7.
8.
9.
~Fx
∃x~Fx
~∃x~Fx
10.
11.
12.
ass id
7 eg
4 r 8 id
6 ud
~∀xFx
2 r 6 id
3 cd
EXERCISES
1. Produce derivations for each of the following (be careful to obey the strategy rule just given):
a. theorem T201: ∴ ∀x(Fx → Gx) → (∀xFx → ∀xGx)
b. half of T204:
∴ ~∃xFx → ∀x~Fx
(similar to the derivation of half of T203)
c. half of theorem T205: ∴ ∀zFx → ~∃x~Fx
Chapter Three -- 28
Version of Aug 2013
CHAPTER 3 SECTION 9
8 SOME DERIVATIONS
Many derivations take a common form. You begin with quantified sentences, and you remove quantifiers.
Then you manipulate formulas using the techniques from chapters 1 and 2. Finally, you restore the
quantifiers. In some cases this is straightforward:
Every bear is friendly
Some bear is dangerous
∴ Something dangerous is friendly
∀x(Px → Qx)
∃y(Py ∧ Ry)
∴ ∃z(Rz ∧ Qz)
First we remove quantifiers using instantiation rules, being careful to apply ei before ui when that is
possible:
1. Show ∃z(Rz ∧ Qz)
2. Pu ∧ Ru
3. Pu → Qu
pr2 ei
pr1 ui
We choose to use 'u' in the universal instantiation step because it gives us something useful. Choosing
other variables or names would be correct, but not useful.
Now we use sentential rules to get a formula that we can existentially quantify:
4.
5.
Qu
Ru ∧ Qu
2 s 3 mp
2 s 4 adj
Now we are in a position to existentially quantify line 5 to get the desired conclusion:
6.
∃z(Rz ∧ Qz)
5 eg
We can then box and cancel:
1. Show ∃z(Rz ∧ Qz)
2.
3.
4.
5.
6.
Pu ∧ Ru
Pu → Qu
Qu
Ru ∧ Qu
∃z(Rz ∧ Qz)
pr2 ei
pr1 ui
2 s 3 mp
2 s 4 adj
5 eg dd
Strategy hint: When a line is available that begins with a universal or existential
quantifier, apply an instantiation rule, ei or ui, to derive an instance.
When the conclusion is a universally quantified formula, it will very likely be derived by using a universal
derivation. When a universal derivation is used, it is usually best to set up that derivation as early as
possible. Consider this example:
Every jaguar is a fast cat
Every cat is an animal
∴ Every jaguar is a fast animal.
∀x(Jx → Fx ∧ Cx)
∀x(Cx → Ax)
∴ ∀x(Jx → Fx ∧ Ax)
Chapter Three -- 29
Version of Aug 2013
CHAPTER 3 SECTION 9
Our initial show line is a universally quantified sentence:
1. Show ∀x(Jx → Fx ∧ Ax)
We can derive line 1 if we can show the formula that you get by removing its initial quantifier. So set that
up as a show line:
2.
Show Jx → Fx ∧ Ax
This is a conditional, so try conditional derivation:
3.
Jx
ass cd
The rest of the conditional derivation is relatively straightforward:
4.
5.
6.
7.
8.
Jx → Fx ∧ Cx
Fx ∧ Cx
Cx → Ax
Ax
Fx ∧ Ax
pr1 ui
3 4 mp
pr2 ui
5 s 6 mp
5 s 7 adj
We have derived the consequent of the conditional to be shown; after boxing and canceling we have:
1. Show ∀x(Jx → Fx ∧ Ax)
2.
Show Jx → Fx ∧ Ax
3.
4.
5.
6.
7.
8.
Jx
Jx → Fx ∧ Cx
Fx ∧ Cx
Cx → Ax
Ax
Fx ∧ Ax
ass cd
pr1 ui
3 4 mp
pr2 ui
5 s 6 mp
5 s 7 adj cd
Since line 2 has been shown, we may infer line 1 by universal derivation:
1. Show ∀x(Jx → Fx ∧ Ax)
2.
Show Jx → Fx ∧ Ax
3.
4.
5.
6.
7.
8.
Jx
Jx → Fx ∧ Cx
Fx ∧ Cx
Cx → Ax
Ax
Fx ∧ Ax
9.
ass cd
pr1 ui
3 4 mp
pr2 ui
5 s 6 mp
5 s 7 adj cd
2 ud
When the conclusion has both universal and existential quantifiers, the strategy is essentially to combine
those above, applying whichever strategy is relevant at the time. Consider this argument:
For every giraffe, there is a leopard which is happy if and only if it (the giraffe) is.
For every leopard, there is a monkey that is happy if and only if it (the leopard) is.
∴ For every giraffe, there is a monkey which is happy if and only if it (the giraffe) is.
∀x(Gx → ∃y(Ly ∧ (Hy ↔ Hx)))
∀x(Lx → ∃y(My ∧ (Hy ↔ Hx)))
∴ ∀x(Gx → ∃y(My ∧ (Hy ↔ Hx)))
The conclusion to be shown is universally quantified, so set up a universal derivation. In fact, this should
generally be done as early as possible.
Chapter Three -- 30
Version of Aug 2013
CHAPTER 3 SECTION 9
Strategy hint: If a universal derivation is to be used to show a universally quantified
formula, ∀x□, set it up as early as possible, by inserting a Show line with ∀x□, and
then proceed to derive the part following the quantifier, namely □.
It is often convenient to immediately follow the show line containing ∀x□ by another containing □.
This is done in line 2 here:
1. Show ∀x(Gx → ∃y(My ∧ (Hy ↔ Hx)))
2.
Show Gx → ∃y(My ∧ (Hy ↔ Hx))
Line 2 is a conditional, so try conditional derivation:
3.
Gx
ass cd
Universally instantiating the first premise and using modus ponens is a natural thing to try:
4.
Gx → ∃y(Ly ∧ (Hy ↔ Hx))
pr1 ui
5.
∃y(Ly ∧ (Hy ↔ Hx))
3 4 mp
We now have derived an existentially quantified formula, and there are some universally quantified ones in
the premises. Generally, when both rules ei and ui are possible, as we stated above, you should use rule
ei first. This is because rule ei introduces a variable which must be brand new in the derivation. If you do
ei first, then you can do ui using the variable introduced by ei. But if you do ui first, you cannot do ei using
that variable. In our derivation, the "ei before ui" strategy is relevant. Apply ei to line 5 using a variable
that does not already occur in the derivation:
6.
Lz ∧ (Hz ↔ Hx)
We can now make use of our second premise to get:
7.
Lz → ∃y(My ∧ (Hy ↔ Hz))
pr2 ui
We can obviously use line 6 to get the consequent of line 7. That consequent is also existentially
quantified, so we apply ei:
8.
9.
∃y(My ∧ (Hy ↔ Hz)
Mu ∧ (Hu ↔ Hz)
6 s 7 mp
8 ei
Now look over what we have and what we want. We are in a conditional derivation, and we need to show
'∃y(My ∧ (Hy ↔ Hx))' to complete that derivation. This formula is existentially quantified, and so we will
probably derive it by existentially generalizing something. That is, we will existentially generalize
something of the form:
M_ ∧ (H_ ↔ Hx)
We already have something very close to that, on line 9; we could get what we want by deriving a formula
just like line 9 but with 'x" instead of 'z'. So suppose we try to derive ' Mu ∧ (Hu ↔ Hx)'. We already have
the left conjunct, so the job is to derive the right conjunct 'Hu ↔ Hx'. This is a biconditional, so we need to
derive two conditionals, probably by conditional derivation, and then put them together by cb. That in fact
is easy to do:
10.
11.
12.
13.
14.
Show Hu → Hx
Hu
Hz
Hx
ass cd
9 s bc 11 mp
6 s bc 12 mp cd
Show Hx → Hu
15.
16.
17.
Hx
Hz
Hu
18.
Hu ↔ Hx
ass cd
6 s bc 15 mp
9 s bc 16 mp cd
10 14 cb
Chapter Three -- 31
Version of Aug 2013
CHAPTER 3 SECTION 9
To finish, we only need to put line 18 together with the first conjunct on line 9, and existentially generalize:
19.
20.
Mu ∧ (Hu ↔ Hx)
∃y(My ∧ (Hy ↔ Hx))
9 s 18 adj
19 eg
This completes our conditional derivation, so we now have:
1. Show ∀x(Gx → ∃y(My ∧ (Hy ↔ Hx)))
2.
Show Gx → ∃y(My ∧ (Hy ↔ Hx))
3.
4.
5.
6.
7.
8.
9.
10.
Gx
Gx → ∃y(Ly ∧ (Hy ↔ Hx))
∃y(Ly ∧ (Hy ↔ Hx))
Lz ∧ (Hz ↔ Hx)
Lz → ∃y(My ∧ (Hy ↔ Hz))
∃y(My ∧ (Hy ↔ Hz)
Mu ∧ (Hu ↔ Hz)
Show Hu → Hx
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
ass cd
pr1 ui
3 4 mp
5 ei
pr2 ui
6 s 7 mp
8 ei
Hu
Hz
Hx
ass cd
9 s bc 11 mp
6 s bc 12 mp cd
Show Hx → Hu
Hx
Hz
Hu
ass cd
6 s bc 15 mp
9 s bc 16 mp cd
Hu ↔ Hx
Mu ∧ (Hu ↔ Hx)
∃y(My ∧ (Hy ↔ Hx))
10 14 cb
9 s 18 adj
19 eg cd
Line 2 has now been shown by the conditional derivation. Now we only need to add line 21, and box and
cancel, finishing the universal derivation.
1. Show ∀x(Gx → ∃y(My ∧ (Hy ↔ Hx)))
2.
Show Gx → ∃y(My ∧ (Hy ↔ Hx))
[[DETAILS ABOVE]]
21.
2 ud
Chapter Three -- 32
Version of Aug 2013
CHAPTER 3 SECTION 9
EXERCISES
1. Symbolize these arguments and provide derivations to validate them. Give an explicit scheme of
abbreviation for each.
a.
If history is right, then if anyone was strong, Hercules was strong.
Only those who work out are strong, and only those with self-discipline work out.
∴ If Hercules does not have self-discipline, then either history is not right or nobody is strong.
b.
If some giraffes are not happy, then all giraffes are morose.
Some giraffes ponder the mysteries of life.
∴ If some giraffes are not morose, then some who ponder the mysteries of life are happy.
c.
There is not a single critic who either likes art or can paint.
Some level-headed people are critics.
Anyone who can't paint is uneducated.
∴ Some level-headed people are uneducated.
d.
No astronaut is a good dancer.
Every singer is warm-blooded.
If something is warm-blooded and is not a good dancer, then nothing that is either a singer or
an astronaut is exultant.
∴ If some astronaut is a singer, then no singer is exultant.
e.
All students who have a sense of humor or are brilliant seek fame.
Anyone who seeks fame and is brilliant is insecure.
Whoever is a mogul is brilliant.
∴ Every student who is a mogul is insecure.
f.
There is a monkey that is happy if and only if some giraffe is happy.
There is a monkey that is happy if and only if some giraffe is not happy.
All monkeys are happy.
∴ It is not the case that either every giraffe is happy or none are.
g.
For every astronaut that writes poetry, there is one that doesn't.
For every astronaut that doesn't write poetry, there is one that does.
∴ If there are any astronauts, some write poetry and some don't.
2. Derive theorems 203-208, 231-232.
T203
T204
T205
T206
T207
T208
T209
T210
~∀xFx ↔ ∃x~Fx
~∃xFx ↔ ∀x~Fx
∀xFx ↔ ~∃x~Fx
∃xFx ↔ ~∀x~Fx
∃x(Fx ∨ Gx) ↔ ∃xFx ∨ ∃xGx
∀x(Fx ∧ Gx) ↔ ∀xFx ∧ ∀xGx
∃x(Fx ∧ Gx) → ∃xFx ∧ ∃xGx
∀xFx ∨ ∀xGx → ∀x(Fx ∨ Gx)
::::::::
T231
T232
∀xFx ↔ ∀yFy
∃xFx ↔ ∃yFy
Chapter Three -- 33
Version of Aug 2013
CHAPTER 3 SECTION 9
9 DERIVED RULES
We have looked at formulas that have quantifiers on their front, or quantifiers that end up on front after a
step such as modus ponens. Things are different if those quantifiers are preceded by a negation sign.
Consider the following simple derivation:
Every A is B.
Nothing is both B and C.
So every A isn't C.
∀x(Ax → Bx)
~∃x(Bx ∧ Cx)
∴ ∀x(Ax → ~Cx)
This is intuitively valid, but deriving it requires slightly indirect reasoning. Our conclusion is universally
quantified, so we set up a universal derivation right away:
1. Show ∀x(Ax → ~Cx)
2. Show Ax → ~Cx
This is a conditional, so we try conditional derivation:
3.
Ax
ass cd
We can spell out some obvious consequences of what we have by instantiating the first premise and
doing modus ponens:
4.
5.
Ax → Bx
Bx
pr1 ui
3 4 mp
The second premise in fact is not anything we can make use of by applying any of our quantifier rules.
Some other approach is needed. At this point it is useful to fall back on a technique from chapter 1; we
are trying to derive '~Cx", so try to derive it by indirect derivation:
6.
7.
Show ~Cx
Cx
ass id
We are not in a position to use the second premise directly, but we can use it indirectly by deriving
something that contradicts it. This is simple in two lines:
8.
9.
Bx ∧ Cx
∃x(Bx ∧ Cx)
5 7 adj
8 eg
Now we complete our indirect derivation with:
10.
~∃x(Bx ∧ Cx)
pr2 9 id
boxing and canceling to get:
1. Show ∀x(Ax → ~Cx)
2. Show Ax → ~Cx
3.
Ax
4.
Ax → Bx
5.
Bx
6.
Show ~Cx
7.
8.
9.
10.
Cx
Bx ∧ Cx
∃x(Bx ∧ Cx)
~∃x(Bx ∧ Cx)
ass cd
pr1 ui
3 4 mp
ass id
5 7 adj
8 eg
pr2 9 id
Chapter Three -- 34
Version of Aug 2013
CHAPTER 3 SECTION 9
This essentially completes the derivation. For line 6 has completed the conditional derivation that starts
on line 2, and once the 'show' on line 2 is cancelled, line 1 follows by universal derivation:
1. Show ∀x(Ax → ~Cx)
2.
3.
4.
5.
6.
Show Ax → ~Cx
Ax
Ax → Bx
Bx
Show ~Cx
ass cd
pr1 ui
3 4 mp
7.
8.
9.
10.
Cx
Bx ∧ Cx
∃x(Bx ∧ Cx)
~∃x(Bx ∧ Cx)
ass id
5 7 adj
8 eg
pr2 9 id
11.
6 cd
12.
2 ud
This kind of indirect strategy is typical of how to handle derivations with sentences that begin with negated
quantifiers when we use only our basic rules for quantifiers. However, it is usually more useful to use
some derived rules that let us replace initial negated quantifiers by unnegated ones of the opposite sort,
which may be used directly. The rule called quantifier negation does this. It lets you replace a negated
initial quantifier by the opposite quantifier followed by a negation. If we lump in all applications of double
negation, we get eight cases:
Rule qn (Quantifier negation)
~∀xFx
∴ ∃x~Fx
~∃xFx
∴ ∀x~Fx
∀xFx
∴ ~∃x~Fx
∃xFx
∴ ~∀x~Fx
~∀x~Fx
∴ ∃xFx
~∃x~Fx
∴ ∀xFx
∀x~Fx
∴ ~∃xFx
∃x~Fx
∴ ~∀xFx
These derived rules are based on T203-206, which are given in the last set of exercises.
Here is how we can use rule qn to shorten the derivation above. We begin as before:
∀x(Ax → Bx)
~∃x(Bx ∧ Cx)
∴ ∀x(Ax → ~Cx)
1. Show ∀x(Ax → ~Cx)
2. Show Ax → Cx
3.
Ax
4.
Ax → Bx
5.
Bx
ass cd
pr1 ui
3 4 mp
Now instead of introducing a subderivation to make indirect use of the second premise, we apply rule qn
to that premise and then make direct use of the result; this lets us proceed quickly to get the desired '~Cx':
Chapter Three -- 35
Version of Aug 2013
CHAPTER 3 SECTION 9
1. Show ∀x(Ax → ~Cx)
2.
Show Ax → ~Cx
3.
4.
5.
6.
7.
8.
9.
Ax
Ax → Bx
Bx
∀x~(Bx ∧ Cx)
~(Bx ∧ Cx)
~Bx ∨ ~Cx
~Cx
ass cd
pr1 ui
3 4 mp
pr2 qn

8 ui
9 dm
5 dn 8 mtp cd
10.
2 ud
The advantage is not just that the derivation is two lines shorter, but the reasoning is simpler, and it is
easier to think up. For that reason we have this strategy hint:
Strategy hint: If an available formula begins with a negation sign immediately followed
by a quantifier which has scope over the rest of the formula, convert it to a more useful
formula by applying rule qn to it.
Here is another example of the use of rule qn. We are given this argument to validate:
~∃x(Ax ∧ Bx)
∀y(Ay ↔ ~Cy)
∀y(Dy → By)
~∀xCx
∴ ∃x~Dx
Neither the first nor the fourth premise may be used as an input to one of the basic quantifier rules.
However, rule qn turns them into useful forms.
1. Show ∃x~Dx
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
∃x~Cx
~Ck
Ak ↔ ~Ck
Ak
∀x~(Ax ∧ Bx)
~(Ak ∧ Bk)
~Ak ∨ ~Bk
~Bk
Dk → Bk
~Dk
∃x~Dx
pr4 qn 
2 ei
pr2 ui
4 bc 3 mp
pr1 qn 
6 ui
7 dm
5 dn 8 mtp
pr3 ui
9 10 mt
11 eg dd
Chapter Three -- 36
Version of Aug 2013
CHAPTER 3 SECTION 9
Rule av: There is another useful derived rule, though one not so often used. Given our explanation of
quantifiers, our choice of bound variables is irrelevant; one is as good as another. This is made explicit in
derived rule av ("alphabetic variance"). The rule says that alphabetically varying the choice of a bound
variable used with an initial quantifier yields an equivalent formula. In particular:
Rule av (alphabetic variance)
From a formula of the form '∀x . . . x . . x . . .', where the initial quantifier has scope over
the whole formula, you may infer '∀y . . . y . . y . . .', which is the result of changing the
variable 'x' in the quantifier to another variable, 'y', and changing all variables inside the
first formula that are bound by the initial quantifier to 'y'.
Likewise if the initial quantifier is '∃' instead of '∀'.
Constraint: No capturing is allowed. That is, this inference is not permitted if a new
occurrence becomes bound by a quantifier inside of the original formula.
As an example, from
∀z(Dz ∧ Ez → ∃u(Du ∨ Fz))
you may infer
∀w(Dw ∧ Ew → ∃u(Du ∨ Fw)).
But you may not infer
∀u(Du ∧ Eu → ∃u(Du ∨ Fu))
because that violates the no capturing rule.
Rule av is based on theorems T231 and T232, proved in the exercises.
Here is a situation in which rule av is useful. Suppose you are given the argument:
∀z(Dz ∧ Ez → ∃u(Du ∨ Fz))
∀x(Dx → ~Fx)
∴ ∀u(Du → ~Eu)
A natural derivation might go like this. The conclusion is universally quantified, so set up a universal
derivation:
1. Show ∀u(Du → ~Eu)
2. Show Du → ~Eu
This is a conditional, so set up a conditional derivation:
3.
Du
ass cd
You now need to show ~Eu, and it is natural to set up an indirect derivation to show this:
4.
5.
Show ~Eu
Eu
ass id
Now universally instantiate the first premise:
6.
Du ∧ Eu → ∃u(Du ∧ Fu)
pr1 ui
Oops, you can't do that! The 'u' following the 'F' gets captured by the quantifier in the consequent of the
conditional. So what can we do? Different ideas might be tried, but it is easy if we use rule av. Don't start
out to derive the conclusion, because it uses a variable that gets you in trouble. Instead, derive a
Chapter Three -- 37
Version of Aug 2013
CHAPTER 3 SECTION 9
sentence that is exactly like the conclusion, but one that uses a different variable. Then use rule av to
change this into the desired conclusion.
Here is a derivation which reaches a sentence just like the conclusion except for using a different variable:
1. Show ∀u(Du → ~Eu)
2.
Show ∀w(Dw → ~Ew)
3.
Show Dw → ~Ew
4.
5.
6.
7.
8.
9.
10.
11.
12.
Dw
ass cd
Show ~Ew
Ew
ass id
Dw ∧ Ew → ∃u(Du ∧ Fw)
pr1 ui
∃u(Du ∧ Fw)
4 6 adj 7 mp
Ds ∧ Fw
8 ei
Fw
9s
Dw → ~Fw
pr2 ui
~Fw
4 11 mp 10 id
13.
 no capturing occurs here
3 ud
Because we used 'x' instead of 'u', we did not encounter any capturing problems in applying rule ui. Now
we merely apply rule av to line 2, and we are done:
1. Show ∀u(Du → ~Eu)
2.
Show ∀w(Dw → ~Ew)
3.
Show Dw → ~Ew
4.
5.
6.
7.
8.
9.
10.
11.
12.
Dw
ass cd
Show ~Ew
Ew
ass id
Dw ∧ Ew → ∃u(Du ∧ Fw)
pr1 ui
∃u(Du ∧ Fw)
4 6 adj 7 mp
Ds ∧ Fw
8 ei
Fw
9s
Dw → ~Fw
pr2 ui
~Fw
4 11 mp 10 id
13.
3 ud
14. ∀u(Du → ~Eu)
2 av dd
Chapter Three -- 38
Version of Aug 2013
CHAPTER 3 SECTION 9
EXERCISES
1. Provide derivations for these arguments.
a.
~∃x(Ax ∨ Bx)
∀x∀y(Gx ∧ Hy → By)
∃xGx
∴ ∀x~Hx
b.
∃x(Hx ∧ ~∃y(Gy ∧ Hx))
∴ ∀y~Gy
c.
∀x(Ax → ∀y(Bx ↔ By))
∃zBz
∴ ∀y(Ay → By)
d.
~∀x(Dx ∨ Ex)
∃x(Fx ↔ ~Ex) → ∀zDz
∴ ∃x~Fx
e.
Jc ∧ ~Jd
∀xKx ∨ ∀x~Kx
∃x(Jx ∧ Kx) → ∀x(Kx → Jx)
∴ ~Kc
2. Provide derivations for these theorems:
T229 ∃x(∃xFx → Fx)
T230 ∃x(Fx → ∃xFx)
T234 ∀x((Fx → Gx) ∧ (Gx → Hx) → (Fx → Hx))
T235 ∀x(Fx → Gx) ∧ ∀x(Gx → Hx) → ∀x(Fx → Hx)
T236 ∀x(Fx ↔ Gx) ∧ ∀x(Gx ↔ Hx) → ∀x(Fx ↔ Hx)
T237 ∀x(Fx → Gx) ∧ ∀x(Fx → Hx) → ∀x(Fx → Gx ∧ Hx)
T238 ∀xFx → ∃xFx
T242 ~∀x(Fx → Gx) ↔ ∃x(Fx ∧ ~Gx)
T243 ~∃x(Fx ∧ Gx) ↔ ∀x(Fx → ~Gx)
T248 ∃xFx ∧ ∃x~Fx ↔ ∀x∃y(Fx ↔ ~Fy)
Chapter Three -- 39
Version of Aug 2013
CHAPTER 3 SECTION 10
10 INVALIDITIES
In chapters 1 and 2 we studied tautological validity, which is formal validity that is due to how sentences
are built up out of sentential letters and connectives. An argument which is tautologically valid is definitely
valid. However, an argument may be valid even if it is not tautologically valid if its validity is due to
something in addition to how it is built up with connectives. We have seen examples of such arguments in
this chapter, arguments such as:
∀xFx
∴ Fa
In this chapter we have studied the kind of formal validity which is due to how formulas are built up out of
names, monadic (one-place) predicates, variables, connectives and quantifiers. We call such validity
"MPC validity" ("monadic predicate calculus validity"). Derivations using the methods of chapters 1-3
show that the arguments they validate are MPC valid. An argument which is MPC valid is definitely valid.
(An argument may be valid even if it is not MPC valid if its validity is due to something in addition to how it
is built up from names, variables, monadic predicates, quantifiers, and connectives. Some examples of
this are:
Some boy fed every cat
<Uses the two-place predicate 'fed'>
∴ Every cat was fed by a boy
There are infinitely many prime numbers
∴ There is at least one prime number.
<Uses the quantifier 'infinitely many'>
Dr. Jekyll is tall
Dr. Jekyll is Mr. Hyde
∴ Mr. Hyde is tall
<Uses 'is' in the sense of identity>
Even though MPC validity is not the whole story, it remains an important kind of validity.)
So far in this chapter we have learned how to show that arguments are MPC valid by means of giving
derivations which validate the arguments. We have not yet focused on how to show that an argument is
not MPC valid. To do that we may describe a logically possible situation in which the argument has true
premises and a false conclusion. It is convenient in doing this to consider very "small" situations -- that is,
situations in which only a small number of things exist. To illustrate this, suppose we are given this
argument:
There are some fibers
Every fiber is green
Something isn't green
∴ Everything green is a fiber
Its MPC form is:
∃xFx
∀x(Fx → Gx)
∃x~Gx
∴ ∀x(Gx → Fx)
Now consider the following "small" situation:
There are three things:
The first is a fiber; the others are not.
The first and the second are green; the third is not.
In this situation the first premise, '∃xFx', is true because the first thing is a fiber. The second premise,
'∀x(Fx → Gx)', is true because there is only one fiber, and it is green. The third premise is true because
something isn't green (the third thing). The conclusion is false because not everything that is green is a
Chapter Three -- 40
Version of Aug 2013
CHAPTER 3 SECTION 10
fiber -- the second thing is green but not a fiber. So this is a situation in which the argument has true
premises and a false conclusion, and so it is not MPC valid.
If we reflect on this technique, we see that all that we need to show that an argument is not MPC valid is
that its logical structure permits this kind of situation. It is sufficient to interpret the argument as applied
to a situation in which there are three things, where we interpret ‘F’ as holding of the first (and no other),
and we interpret ‘G’ as holding of the first and second (but not the third). If you can interprert an argument
so that its premises come out true when so interpreted, and its conclusion false, this is enough to show
that an argument of the given form isn't MPC valid. We call such an interpretation a counter-example for
the argument. We can describe such a counter-example by using a format like this:
Universe:
First thing Second thing Third thing
F: {the first thing}
G: {the first thing, the second thing}
This indicates how many things there are in the situation, and it gives the "extensions" of 'F' and 'G'. The
extension of a predicate is just the set of things it is true of in that interpretation. So the information above
tells us that 'F' is true of the first thing and of nothing else, and it tells us that 'G' is true of the first and
second things, and not of the third.
It doesn’t matter at all what these things are. For specificity, it is often convenient to use integers:
Universe:
0 1 2
F: {0}
G: {0, 1}
This information describes a counter-example for the original argument, because it describes, in minimal
terms, the structure of a situation in which the premises of the argument are true and the conclusion false.
Here are some more arguments that are not MPC valid, and counter-examples for them.
Counter-example #2:
∃x(Fx ∧ ~Gx)
∀x(Hx → ~Gx)
∃x(Hx ∧ Fx)
∴ ∀x(Fx → ~Gx)
Universe:
0 1 2
F: {0, 1}
G: {0, 2}
H: {1}
The first premise is true in this interpretation because 'F' is true of 1 and G isn't. The second premise is
true because everything that 'H' is true of, namely 1, 'G' is not true of, and the third premise is true
because both 'F' and 'H' are true of 1. But the conclusion is not true, because not everything that 'F' is true
of is something that 'G' is not true of; 0 is an example. (Removing 2 from the universe will also yield a
counter-example.)
Chapter Three -- 41
Version of Aug 2013
CHAPTER 3 SECTION 10
If the argument contains name letters, we indicate what they stand for in the given universe:
Counter-example #3:
∀x(Ax → (Bx ↔ Cx))
Bk ∧ ~Ck
∴ ∀x~Ax
Universe:
0 1
A: {1}
B: {0, 1}
C: {1}
k: 0
<'k' stands for the first thing>
The first premise is true because whatever 'A' is true of, namely 1, is such that 'B' and 'C' are both true of
it, so their biconditional comes out true. The second premise is true because 'B' is true of what 'k' stands
for, namely 0, and 'C' isn't. The conclusion is false because 'A' is not false of everything; it is true of 1.
Counter-example #4:
∀x∃y(Ax ↔ By)
∃xBx ∧ ∃x~Bx
∀x(Ax → ~Cx)
∴ ~∀xCx
Universe:
0 1
A: { }
B: {0}
C: {0, 1}
<true of nothing at all>
The first premise is true because everything is such that something is such that 'A' is true of the first if and
only if 'B' is true of the second. In fact, 'A' is true of nothing at all. And no matter what there is, there is
something that 'B' is not true of, namely 1. So there is always something that makes the biconditional
true. The second premise is clearly true since 'B' is true of something, namely 0, and 'B' is also false of
something, namely 1. The third premise is true because 'A' is true of nothing, so that every instance is a
conditional with a false antecedent. The conclusion is false because 'C' is indeed true of everything.
Thinking up counter-examples: If you believe that an argument is not MPC valid, how do you think up a
counter-example? There is a mechanical way to do this (described below), but it is too complex to be
useful in many cases. So we will usually have to be creative. Some general observations may be useful
in guiding our creativity. One approach that is often used is to build up the counter-example a piece at a
time, guided by what is needed to make the premises true and conclusion false. Suppose we are given
this argument:
∀x~(Fx ↔ Hx)
∃x(Hx ∧ Gx)
∃x(Hx ∧ ~Gx)
∴ ∀x(Fx → Gx)
So far, we don't know what will be in the universe. Begin by asking what is needed to make the
conclusion false. In this case, what is needed is that there be something that 'F' is true of and 'G' is not.
So write this:
F: {0}
G: { }
<not 0>
The notation "<not 0>" at the right is not part of the counter-example; it is merely a reminder to yourself
that when constructing the counter-example you should not add 0 to the list of things that 'G' is true of,
because that could make the conclusion true.
Chapter Three -- 42
Version of Aug 2013
CHAPTER 3 SECTION 10
Now consider the first premise; this says that whatever there is in the universe, 'F' and 'H' must disagree
about it. This must be kept in mind as a constraint on what can be in the counter-example. So far, in fact,
it tells us that since 'F' is true of 0, 'H' must not be:
F: {0}
G: { }
H: { }
<not 0>
<not 0>
Next, consider the second premise: 'G' and 'H' are both true of something. It can't be 0, so fill in 1:
F: {0}
G: {1}
H: {1}
<not 0>
<not 0>
Next, the third premise; this says that there is something that 'H' is true of which 'G' is not true of. It can't
be 0 because 'H' cannot be true of 0. It can't be 1 because 'G' is true of 1. So there must be a third thing:
F: {0}
G: {1}
<not 0>
H: {1, 2}
<not 0>
At this point we have all of the information we need. This is our proposed counter-example:
Universe:
0 1 2
F: {0}
G: {1}
H: {1, 2}
If you check through the parts of the argument, you will see that the premises are all true and the
conclusion false.
Sometimes if you start with no predicate being true of anything, a counter-example falls into your lap.
Here is such a case. The argument is:
∀x(Jx → Kx ∨ Hx)
~∀x(~Kx → Jx)
~∃x(Kx ∧ ~Hx)
Hc → ∃xJx
∴ ~∃x(Hx ∨ ~Jx)
Begin with this minimal proposed counter-example:
Universe:
0
H: { }
J: { }
K: { }
c: 0
Let us see what we need to add to what the predicates are true of to make this a counter-example. The
first premise is already true because it is a quantified conditional with an antecedent that is false for each
thing in the universe. The second premise is true because its unnegation '∀x(~Kx → Jx)' is false. This is
false because the part following the quantifier: '~Kx → Jx' is not true for every way of treating 'x' like a
name; it is false when 'x' stands for 0. The third is true because there is nothing that is K. The fourth is
true because it is a conditional with a false antecedent. And the conclusion is false because there is
indeed something that is either H or not J; 0 is not J, so it is either H or not J. In short, the counterexample works as stated. (Usually, of course, more work will be needed.)
Chapter Three -- 43
Version of Aug 2013
CHAPTER 3 SECTION 10
You may sometimes wonder how many things to put into the universe in order to produce a counterexample. There is no best way to determine this; usually you just put more things in when that seems to
be required by the premises being true and the conclusion false. There is, however, an upper limit on
what you need. If there is only one predicate letter in the argument, then you will need no more than two
things in the universe. If there are two predicate letters, you will need no more than four. If there are three
predicate letters, you will need no more than eight things. And so on. There is a formula for this: if there
n
are n predicate letters, if there is a counter-example, there is one using no more than 2 things.
Name letters have no effect on the number of things needed. If there are only two predicate letters, and
thirteen name letters, then if there is a counter-example at all, there is one with four or fewer things.
(Of course, if there are four things and thirteen name letters, several different names will have to stand for
the same things. But that's OK.)
So here is a mechanical way to come up with a counter-example. Decide, by the formula above, the
maximum number of things needed in the universe for a counter-example. For example, suppose that
2
there are two monadic predicates in the argument. A universe of size 2 , that is, 4, will do. Now just
consider what choices there may be for the extension of predicate 'F'. There are 16 options:
{ }, {0}, {1}, {2}, {3}, {0, 1}, {0, 2}, {0, 3}, {1, 2}, {1,3}, {2,3}, {0, 1, 2}, {0, 1, 3},
{0, 2, 3}, {1, 2, 3}, {0, 1, 2, 3}
There are also 16 options for 'G'. So there are 16×16 = 256 options for possible counter-examples. If you
just check these out, one at a time, you are sure to find one if one exists. If there are three monadic
predicates there are 4,096 options. And so on.
(Exercise for the reader: In the above calculation we have supposed that if there is a counter-example, we
can find one using a maximum size universe. We have ignored the possibility that there is, say, a
counter-example using a universe of size 3 but none using a universe of size 4. Why are we justified in
making that assumption?)
EXERCISES
1. Give counter-examples for each of the following arguments.
a.
∀x(Ax → ∃y(By ∧ ~Ay))
~∀xBx
~∃x(Bx ∧ Cx)
∴ ∃x(Ax ∧ Cx)
b.
∃x(Dx ∧ Ex ∧ ~Fx)
∃x(~Dx ∧ ~Ex)
∀x(Ex → Dx ∨ Fx)
∴ ∀x(Dx ∧ Ex → ~Fx)
c.
∃x(Fx ∧ Gx)
∃x(Fx ∧ ~Gx)
∃x(~Fx ∧ Gx)
∴ ∀x(~Fx → Gx)
d.
∀x∃y(Fx ↔ (Gy ∨ Fx))
∴ ~∃xFx → ~∃xGx
e.
Ha ∧ ~Hb
∀x(Kx → Hx ∧ Jx)
∃x(Jx ∧ ~Kx)
∴ ∃x(Hx ∧ ~Jx)
<requires more than three things in the universe>
Chapter Three -- 44
Version of Aug 2013
CHAPTER 3 SECTION 11
11 EXPANSIONS
In constructing counter-examples it is sometimes difficult to assess the truth value of a sentence in the
counter-example, especially when it contains overlapping quantifiers. For example, ask yourself whether
the following is a legitimate counter-example to this argument:
∀x∃y(Ax ↔ ~Ay)
∃x(Ax ∧ Bx)
∴ ∀xAx
Universe:
0
1
2
A: {0}
B: {0}
It is clear that this makes the conclusion false, and the second premise true. What about the first
premise? It makes that true too. The first premise says that every thing in the universe is such that, there
is a thing in the universe such that it isn't A if the first thing is A, and it is A if the first thing isn't. This is in
fact true in the counter-example. But this may not be obvious to you. If not, there is a mechanical way to
answer such a question. It resembles truth tables in that it will automatically give you a yes or no answer,
but it may involve complexity. The technique is based on the idea that if there are a small number of
things in the universe, then a universally quantified claim is equivalent to a conjunction of unquantified
claims got by removing the quantifier and applying each resulting claim to a thing in the universe. And an
existentially quantified claim, in turn, is equivalent to a disjunction of such claims that are applied to each
thing in the universe.
Let us introduce a convention for naming things in a universe. When there are three things the names will
be 'i0', 'i1', and 'i2', where:
'i0' stands for 0
'i1' stands for 1
'i2' stands for 2.
(If there are fewer things, leave out 'i2', or both 'i1' and 'i2'. If there are more things add 'i4', 'i5', and so on.)
Now consider the sentence '∀xAx'. This says that everything in the universe is A. This is equivalent to
saying that the first thing is A and the second thing is A and the third thing is A. That is, it is equivalent to
the conjunction:
∀xAx
is equivalent to
Ai0 ∧ Ai1 ∧ Ai2
It is easy to check that this conjunction is false, because not all conjuncts are true.
The second premise is '∃x(Ax ∧ Bx)'. This is equivalent to saying that either the first thing is both A and B,
or the second thing is, or the third. That is, the quantified sentence is equivalent to this disjunction:
∃x(Ax ∧ Bx)
is equivalent to
(Ai0 ∧ Bi0) ∨ (Ai1 ∧ Bi1) ∨ (Ai2 ∧ Bi2)
It is easy to check that this disjunction is true, because at least one disjunct is true; the first disjunct is true.
The first premise, '∀x∃y(Ax ↔ ~Ay)', is more interesting. It is universally quantified, so it is equivalent to
the following conjunction:
∃y(Ai0 ↔ ~Ay) ∧ ∃y(Ai1 ↔ ~Ay) ∧ ∃y(Ai2 ↔ ~Ay)
It may be easy to determine that this is true.
The first conjunct is true because there is something which is not A if and only if 0 is A. We know
that 0 is A, and there is indeed at least one thing which is not A; for example, 1 is not A.
The second conjunct is true because there is something which is not A if and only if 1 is A. We
know that 1 is not A, and there is indeed at least one thing which is A; for example, 0 is A.
Chapter Three -- 45
Version of Aug 2013
CHAPTER 3 SECTION 11
The third conjunct is just like the second; it is true because there is something which is not A if
and only if 2 is A. We know that 2 is not A, and there is indeed at least one thing which is A; for
example, 0 is A.
Even this, however, is a bit subtle. There is a way to make it even more mechanical. Namely, each
existentially quantified biconditional is equivalent to a disjunction. So this three-part conjunction:
∃y(Ai0 ↔ ~Ay) ∧
∃y(Ai1 ↔ ~Ay) ∧
∃y(Ai2 ↔ ~Ay)
is equivalent to this:
( (Ai0 ↔ ~Ai0) ∨ (Ai0 ↔ ~Ai1) ∨ (Ai0 ↔ ~Ai2) ) ∧
( (Ai1 ↔ ~Ai0) ∨ (Ai1 ↔ ~Ai1) ∨ (Ai1 ↔ ~Ai2) ) ∧
( (Ai2 ↔ ~Ai0) ∨ (Ai2 ↔ ~Ai1) ∨ (Ai2 ↔ ~Ai2) )
And this is easy to evaluate. There are only three atomic sentences in this complex sentence: 'Ai0', 'Ai1',
and 'Ai2'. The first of these is true, and the others are false. It is thus easy to evaluate the biconditionals:
( (Ai0 ↔ ~Ai0) ∨ (Ai0 ↔ ~Ai1) ∨ (Ai0 ↔ ~Ai2) ) ∧
false
true
true
( (Ai1 ↔ ~Ai0) ∨ (Ai1 ↔ ~Ai1) ∨ (Ai1 ↔ ~Ai2) ) ∧
true
false
false
( (Ai2 ↔ ~Ai0) ∨ (Ai2 ↔ ~Ai1) ∨ (Ai2 ↔ ~Ai2) )
true
false
false
Each disjunction has a true disjunct, so each is true. So the conjunction of the disjunctions is also true.
That is, the whole sentence, which is equivalent to '∀x∃y(Ax ↔ ~Ay)', is true. This process is tedious, but
completely mechanical.
If the counter-example has a universe of only one thing, then this device is very easy to apply. Consider
this argument and the accompanying counter-example:
∀x∀y(Jx ↔ ∃z(Kz ↔ Jy))
∴ ∀xJx
Universe:
0
J: { }
K: {0}
It is clear that the conclusion is false, because 'J' is not true of 0. The premise is universally quantified, so
it is equivalent to a conjunction of all of its instances using names of things in the universe. Since there is
only one thing in the universe, this conjunction has only one conjunct. So:
∀x∀y(Jx ↔ ∃z(Kz ↔ Jy))
is equivalent to
∀y(Ji0 ↔ ∃z(Kz ↔ Jy))
But that in turn has a simpler equivalent:
∀y(Ji0 ↔ ∃z(Kz ↔ Jy))
is equivalent to
Ji0 ↔ ∃z(Kz ↔ Ji0)
In 'Ji0 ↔ ∃z(Kz ↔ Ji0)' the existentially quantified formula on the right is equivalent to a disjunction with
only one disjunct, so we finally have:
Ji0 ↔ (Ki0 ↔ Ji0)
The truth values of the parts of this sentence are:
Chapter Three -- 46
Version of Aug 2013
CHAPTER 3 SECTION 11
'Ji0 ↔ (Ki0 ↔ Ji0)'
false true false
and the whole sentence is true, as desired.
One more example. Consider the argument, and proposed counter-example:
∀x∃y(Fx ∨ Gy)
~∀xFx
~∀xGx
∴ ~∃xGx
Universe:
0
1
F: {0}
G: {1}
It is pretty clear that this proposed counter-example makes the conclusion false, since something is G,
namely, 1. The third premise is true since not everything is G; 0 isn't G. Likewise, the second premise is
true since not everything is F; 1 is not F. What about the first? If you are not certain, you can expand it.
In this proposed counter-example, the sentence '∀x∃y(Fx ∨ Gy)', which starts with a universal quantifier, is
equivalent to this conjunction:
∃y(Fi0 ∨ Gy) ∧ ∃y(Fi1 ∨ Gy)
Each of the existentially quantified sentences is equivalent to a disjunction, so we have:
((Fi0 ∨ Gi0) ∨ (Fi0 ∨ Gi1)) ∧ ((Fi1 ∨ Gi0) ∨ (Fi1 ∨ Gi1))
evaluating the parts we have:
((Fi0 ∨ Gi0) ∨ (Fi0 ∨ Gi1)) ∧ ((Fi1 ∨ Gi0) ∨ (Fi1 ∨ Gi1))
true
false true true
false false false true
true
true
Each conjunct is true, so the sentence is itself true.
Chapter Three -- 47
Version of Aug 2013
CHAPTER 3 SECTION 11
EXERCISES
0. For each of the following arguments use the method of expansions to determine whether the following
is a counterexample for it or not.
Universe:
0
1
2
F: {0}
G: {0, 2}
H: {2}
a.
∀x(Hx → ∃y(Fy ∧ ~Hy))
~∀xFx
~∃x(Fx ∧ Gx)
∴ ∃x(Hx ∧ Gx)
b.
∃x(Gx ∧ Hx ∧ ~Fx)
∃x(~Gx ∧ ~Hx)
∀x(Hx → Gx ∨ Fx)
∴ ∀x(Gx ∧ Hx → ~Fx)
c.
∃x(Fx ∧ Gx)
∃x(Fx ∧ ~Gx)
∃x(~Fx ∧ Gx)
∴ ∀x(~Fx → Gx)
d.
∀x∃y(Fx ↔ (Gy ∨ Fx))
∴ ~∃xFx → ~∃xGx
e.
Ha ∧ ~Hb
∀x(Fx → Hx ∧ Gx)
∃x(Gx ∧ ~Fx)
∴ ∃x(Hx ∧ ~Gx)
Chapter Three -- 48
Version of Aug 2013
CHAPTER 3 RULES
BASIC RULES AND DERIVATION TECHNIQUES FOR CHAPTER 3
Rule ui: (universal instantiation):
∴
∀x ...x...x...
…b…b…
∴
∀x ...x...x...
…y…y…
Every occurrence of 'x' that '∀x' was binding must be replaced with the same name or
variable.
A new variable must not be introduced if some of its new occurrences are bound by a
quantifier in the original formula.
Rule eg (existential generalization):
...b...b...
∴ ∃x…x…b…
...y...y...
∴ ∃x…x…b…
(You need not replace every occurrence of 'b' or of 'y' by 'x'.)
A new variable must not be introduced if some of its new occurrences are bound by a
quantifier in the original formula.
Rule ei: (existential instantiation):
∴
∃x ...x...x...
…y…y…
You must replace every occurrence of 'x' that '∃x' was binding.
The variable 'y' must not already occur in the derivation or in a premise cited in
the derivation..
Universal derivation:
If you have a derivation of the following form:
Show ∀x . . . x . . . x . . .
:::::
:::::
...x...x...
Then if there are no uncancelled show lines in between the first and last lines
displayed, and if 'x' does not occur free anywhere in the derivation that is available
from the show line (or in a premise that has been cited on an available line), you may
box and cancel, using the notation 'ud'.
Chapter Three -- 49
Version of Aug 2013
CHAPTER 3 RULES
DERIVED RULES
Rule qn (Quantifier negation)
~∀xFx
∴ ∃x~Fx
~∃xFx
∴ ∀x~Fx
∀xFx
∴ ~∃x~Fx
∃xFx
∴ ~∀x~Fx
~∀x~Fx
∴ ∃xFx
~∃x~Fx
∴ ∀xFx
∀x~Fx
∴ ~∃xFx
∃x~Fx
∴ ~∀xFx
Rule av (alphabetic variance)
From a formula of the form '∀x . . . x . . x . . .', where the initial quantifier has scope
over the whole formula, you may infer '∀y . . . y . . y . . .', which is the result of
changing the variable 'x' in the quantifier to another variable, 'y', and changing all
variables inside the first formula that are bound by the initial quantifier to 'y'.
Likewise if the initial quantifier is '∃' instead of '∀'.
Constraint: No capturing is allowed. That is, this inference is not permitted if the new
variable becomes bound by a quantifier inside of the original formula.
Chapter Three -- 50
Version of Aug 2013
CHAPTER 3 STRATEGIES
STRATEGY HINTS
All of the strategy hints from chapters 1 and 2 still apply.
These are new:
To derive:
Try this:
Universal Quantification
∀x□
Set up a universal derivation. Write a show line containing ∀x□, and
then immediately follow this with a show line containing □. When the
second show is cancelled, use rule ud to cancel the first.
Or write a show line with '∀x□', and then assume '~∀x□' for an indirect
derivation. Turn this into '∃x~□', and proceed from there.
Existential Quantification
∃x□
Negation of a Universal
Quantification
~∀x□
Negation of an Existential
Quantification
~∃x□
Derive an instance and then use rule eg.
Or write a show line with '∃x□', and then assume '~∃x□' for an indirect
derivation. Turn this into '∀x~□', and proceed from there.
State a show line with '~∀x□', and then assume '∀x□' for an indirect
derivation.
Or derive '∃x~□' and apply derived rule qn.
State a show line with '~∃x□', and then assume '∃x□' for an indirect
derivation.
Or derive '∀x~□' and apply derived rule qn.
If you have this available:
Try this:
Universal Quantification
∀x□
Use rule ui to derive an instance.
(But use rule ei first if that is an option.)
Existential Quantification
∃x□
Use rule ei to derive an instance.
Negation of a Universal
Quantification
~∀x□
Use derived rule qn to turn this into an existential quantification.
Negation of an Existential
Quantification
~∃x□
Use derived rule qn to turn this into a universal quantification.
Use rule av if necessary: If you are having difficulty with capturing when you use rule ui or ei, change
what you are trying to derive to an alphabetic variant. Complete the derivation, and then use derived rule
av to convert this into a derivation of what you are after.
Chapter Three -- 51
Version of Aug 2013
CHAPTER 3 THEOREMS
CHAPTER 3 THEOREMS
LAWS OF DISTRIBUTION:
T201
T202
T207
T208
T209
T210
T211
T212
T213
T214
∀x(Fx → Gx) → (∀xFx → ∀xGx)
∀x(Fx → Gx) → (∃xFx → ∃xGx)
∃x(Fx ∨ Gx) ↔ ∃xFx ∨ ∃xGx
∀x(Fx ∧ Gx) ↔ ∀xFx ∧ ∀xGx
∃x(Fx ∧ Gx) → ∃xFx ∧ ∃xGx
∀xFx ∨ ∀xGx → ∀x(Fx ∨ Gx)
(∃xFx → ∃xGx) → ∃x(Fx → Gx)
(∀xFx → ∀xGx) → ∃x(Fx → Gx)
∀x(Fx ↔ Gx) → (∀xFx ↔ ∀xGx)
∀x(Fx ↔ Gx) → (∃xFx ↔ ∃xGx)
LAWS OF QUANTIFIER NEGATION
T203
T204
T205
T206
~∀xFx ↔ ∃x~Fx
~∃xFx ↔ ∀x~Fx
∀xFx ↔ ~∃x~Fx
∃xFx ↔ ~∀x~Fx
LAWS OF CONFINEMENT
T215
T216
T217
T218
T219
T220
T221
T222
T223
T224
T225
T226
∀x(P∧Fx) ↔ P∧∀xFx
∃x(P∧Fx) ↔ P∧∃xFx
∀x(P∨Fx) ↔ P∨∀xFx
∃x(P∨Fx) ↔ P∨∃xFx
∀x(P→Fx) ↔ (P→∀xFx)
∃x(P→Fx) ↔ (P→∃xFx)
∀x(Fx→P) ↔ (∃xFx→P)
∃x(Fx→P) ↔ (∀xFx→P)
∀x(Fx↔P) → (∀xFx→P)
∀x(Fx↔P) → (∃xFx→P)
(∃xFx↔P) → ∃x(Fx↔P)
(∀xFx↔P) → ∃x(Fx↔P)
LAWS OF VACUOUS QUANTIFICATION
T227
T228
T229
T230
∀xP ↔ P
∃xP ↔ P
∃x(∃xFx → Fx)
∃x(Fx → ∀xFx)
LAWS OF ALPHABETIC VARIANCE
T231
T232
∀xFx ↔ ∀yFy
∃xFx ↔ ∃yFy
Chapter Three -- 52
Version of Aug 2013
CHAPTER 3 THEOREMS
OTHER
T233
T234
T235
T236
T237
T238
T239
T240
T241
T242
T243
T244
T245
T246
T247
T248
(Fx→Gx) ∧ (Gx→Hx) → (Fx→Hx)
∀x((Fx → Gx) ∧ (Gx → Hx) → (Fx → Hx))
∀x(Fx → Gx) ∧ ∀x(Gx → Hx) → ∀x(Fx → Hx)
∀x(Fx ↔ Gx) ∧ ∀x(Gx ↔ Hx) → ∀x(Fx ↔ Hx)
∀x(Fx → Gx) ∧ ∀x(Fx → Hx) → ∀x(Fx → Gx ∧ Hx)
∀xFx → ∃xFx
∀xFx ∧ ∃xGx → ∃x(Fx∧Gx)
∀x(Fx→Gx) ∧ ∃x(Fx∧Hx) → ∃x(Gx∧Hx)
∀x(Fx→Gx∨Hx) → ∀x(Fx →Gx) ∨ ∃x(Fx∧Hx)
~∀x(Fx → Gx) ↔ ∃x(Fx ∧ ~Gx)
~∃x(Fx ∧ Gx) ↔ ∀x(Fx → ~Gx)
~∃xFx → ∀x(Fx→Gx)
~∃xFx ↔ ∀x(Fx→Gx) ∧ ∀x(Fx→~Gx)
~∃xFx ∧ ~∃xGx → ∀x(Fx↔Gx)
∃x(Fx→Gx) ↔ ∃x~Fx ∨ ∃xGx
∃xFx ∧ ∃x~Fx ↔ ∀x∃y(Fx ↔ ~Fy)
Chapter Three -- 53
Version of Aug 2013
Answers to Exercises -- Chapter 3
Answers to the Exercises -- Chapter 3
SECTION 1
1. a. Fred is an orangutan.
Of
b. Gertrude is an orangutan but Fred isn't.
Gertrude is an orangutan [and] Fred is not [an orangutan].
Og ∧ ~Of
c. Tony Blair will speak first.
Fb
d. Gary lost weight recently; he is happy.
Gary lost weight recently [and] [Gary] is happy.
Lg ∧ Hg
e. Felix cleaned and polished.
Felix cleaned and [Felix] polished.
Cf ∧ Of
f. Darlene or Abe will bat clean-up.
Darlene [will bat clean-up] or Abe will bat clean-up.
Bd ∨ Ba
2. 'D' is true of doctors
'L' is true of people who are in love
'h' stands for Hans
'a' stands for Amanda
a.
Hans is a doctor but Amanda isn't.
Hans is a doctor [and] Amanda is not [a doctor]
Dh ∧ ~Da
b.
Hans, who is a doctor, is in love
Hans is in love [and Hans] is a doctor
Lh ∧ Dh
c.
Hans is in love but Amanda isn't
Hans is in love [and] Amanda is [not in love]
Lh ∧ ~La
d.
Neither Hans nor Amanda is in love
[It is not the case that] (Hans [is in love] or Amanda is in love)
~(Lh ∨ La)
f.
Hans and Amanda are both doctors.
Hans is a doctor [and] Amanda is a doctor.
Dh ∧ Da
3. 'L' for things that live in Brea
'D' for things that drive to school
Copyrighted material
Chapter Three -- 54 -- Answers to the Exercises
Version of Aug 2013
Answers to Exercises -- Chapter 3
a.
Eileen and Cosi both live in Brea.
Eileen lives in Brea and Cosi loves in Brea
Le ∧ Lc
b.
Eileen drives to school, and so does Hank.
Eileen drives to school and hank drives to school
De ∧ Dh
c.
If Hank lives in Brea then he drives to school; otherwise he doesn't drive to school.
(If Hank lives in Brea then he drives to school) [and] (otherwise he doesn't drive to school)
(If Hank lives in Brea then he drives to school) [and] ([if Hank doesn't live in Brea then] he doesn't drive to school)
(Lh → Dh) ∧ (~Lh → ~Dh)
d.
If David and Hank both live in Brea then David drives to school but Hank doesn't.
If (David and Hank both live in Brea) then (David drives to school [and] Hank doesn't [drive to school])
(Ld ∧ Lh) → (Dd ∧ ~Dh)
e.
Neither Hank nor Eileen live in Brea, yet each of them drives to school.
Neither Hank nor Eileen live in Brea, [and] [Hank and Eileen] drive to school.
~(Lh ∨ Le) ∧ (Dh ∧ De)
SECTION 2
1. For each of the following, say whether it is a formula in official notation, or in informal notation, or not a
formula at all. If it is a formula, parse it.
a. Official notation
~∀x(Fx → (Gx ∧ Hx))
|
∀x(Fx → (Gx ∧ Hx))
|
(Fx → (Gx ∧ Hx))
/\
Fx (Gx ∧ Hx)
/\
Gx Hx
b. Informal notation
∃x~~Gx → Hx ∨ ∃yGy
/\
∃x~~Gx Hx ∨ ∃yGy
|
/\
~~Gx
Hx ∃yGy
|
|
~Gx
Gy
|
Gx
c. Official notation
~(Gx ↔ ~Hx)
|
(Gx ↔ ~Hx)
Copyrighted material
Chapter Three -- 55 -- Answers to the Exercises
Version of Aug 2013
Answers to Exercises -- Chapter 3
/\
Gx ~Hx
|
Hx
d. Not a formula; a quantifier cannot occur outside a quantifier phrase.
e. Informal notation
Fa → (Gb ↔ Hc)
/\
Fa (Gb ↔ Hc)
/\
Gb Hc
f. Not a formula; a variable can only occur in an atomic formula or a quantifier phrase, and never by itself.
g. Informal notation
∀x(Gx ↔ Hx) → Ha ∧ ∃zKz
/\
∀x(Gx ↔ Hx) Ha ∧ ∃zKz
|
/\
Gx ↔ Hx
Ha ∃zKz
/\
|
Gx Hx
Kz
SECTION 3
1. a.
Sentence
∃x(Fx ∧ ∀y(Gy ∨ Hx))
b.
c.
Not a formula; there is no way to form "∃~z" in our grammar.
Formula
∃z~(Hz ∧ Gx ∧ ∃xIx)
d.
Formula
~(~Gx → ∀y(Jx ∧ Ky ↔ Lx))
e.
Formula
∃xGx ↔ ∃y(Gy ∧ Hx)
f.
Sentence
∀x(Gx → ∀y(Hy → ∀z ( Iz → Hx ∧ Gz)))
g.
Sentence
∀x ∃y( Hx ↔ ~Gy)
h.
i.
j.
Not a formula; there is no way to form "∀xy" in our grammar.
Not a formula; "∃y" cannot stand on its own as a subformula.
Sentence
∀x ∃y∀z(Gx ↔ ∃w(Hw ∧ ~Hx ∧ Gy))
Copyrighted material
Chapter Three -- 56 -- Answers to the Exercises
Version of Aug 2013
Answers to Exercises -- Chapter 3
SECTION 4
1. a. Something is a sofa and is well built. There is a well-built sofa..
b. Everything is such that if it is a sofa then it is well-built. All sofas are well-built.
c. Everything is either a sofa or is well-built. Everything is a sofa, unless it's well-built.
d. Something is such that it is not a sofa. Something isn't a sofa.
e. Everything is such that it is not a sofa. There are no sofas.
f. Everything is such that if it is both bell-built and a sofa, then it is comfortable. Every well-built sofa is
comfortable.
g. Something is comfortable and everything is well-built.
h. Something is such that if it is comfortable, then everything is well-built.
2. Assume that all giraffes are friendly, and that some giraffes are clever and some aren't.
a. ∀x(Gx → Fx)
b. ∀x(Gx → Cx)
c. ∃x(~Fx ∧ Gx)
d. ∃y(Fy ∧ Cy)
e. ∃z(Gz ∧ Cz)
f. ∀x(Gx → ~Gx)
True, since all giraffes are friendly.
False, since not every giraffe is clever.
False, since every giraffe is friendly.
True, since giraffes are friendly, and some of them are clever.
True, since some giraffes are clever.
False, since not every giraffe isn't a giraffe. (In fact, no giraffe isn't a giraffe, but it
only takes one to falsify the symbolic sentence.)
SECTION 5a
1. a.
b.
c.
d.
e.
f.
g.
Every Handsome Elephant is Friendly.
∀x((Hx ∧ Ex) → Fx)
No handsome elephant is friendly.
~∃x((Hx ∧ Ex) ∧ Fx)
Some elephants are not handsome.
∃x(Ex ∧ ~Hx)
Some handsome elephants are friendly.
∃x((Hx ∧ Ex) ∧ Fx)
Each friendly elephant is handsome.
∀x((Fx ∧ Ex) → Hx)
A handsome elephant is not friendly.
∃x((Hx ∧ Ex) ∧ ~Fx)
No friendly elephant is handsome.
~∃x((Fx ∧ Ex) ∧ Hx)
SECTION 5b
1. Suppose that `A' stands for `is a U.S. state', `C' for `is a city', `L' for `is a capital', and `E' for `is in the
Eastern time zone'. What are the truth values of these sentences?
a.
b.
c.
d.
e.
f.
g.
h.
∀x(Cx → Lx) --- False; Los Angeles is a city but not a capital.
∃x(Cx ∧ Lx) --- True; Sacramento is a city and a capital.
∃x(Cx ∧ Lx ↔ Ex) --- True, because something makes the biconditional true, by making both
sides false. For example, Los Angeles is not a capital, and it is not in the Eastern time zone.
∀x(Cx ∧ Ex → Ax) --- False; Philadelphia is not a state.
~∃x(Ax ∧ Ex) --- False; Delaware is a state in the Eastern time zone.
∃x(Cx ∧ Ex) ∧ ∃x(Cx ∧ ~Ex) --- True; Philadelphia is a city in the Eastern time zone and LA is a
city outside the eastern time zone.
∃x(Cx ∧ Ex ∧ Ax) --- False; no city is also a state.
~∃x(Cx ∧ ~Cx) --- True. There is no city which isn't a city.
Copyrighted material
Chapter Three -- 57 -- Answers to the Exercises
Version of Aug 2013
Answers to Exercises -- Chapter 3
2. a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.
m.
n.
o.
All Giraffes are spOtted.
∀x(Gx → Ox)
All Clever giraffes are spotted.
∀x(Gx ∧ Cx → Ox)
No clever giraffes are spotted.
~∃x(Gx ∧ Cx ∧ Ox)
Every giraffe is either spotted or Drab.
∀x(Gx → (Ox ∨ Dx))
Some giraffes are clever.
∃x(Gx ∧ Cx)
Some spotted giraffes are clever.
∃x(Ox ∧ Gx ∧ Cx)
Some giraffes are clever and some aren't.
Some giraffes are clever and some [giraffes are not clever].
∃x(Gx ∧ Cx) ∧ ∃x(Gx ∧ ~Cx)
Some spotted giraffes aren't clever.
∃x(Ox ∧ Gx ∧ ~Cx)
No spotted giraffe is clever but every unspotted one is.
No spotted giraffe is clever [and] every un-spotted [giraffe] is [clever].
~∃x(Ox ∧ Gx ∧ Cx) ∧ ∀x(~Ox ∧ Gx → Cx)
Every clever spotted giraffe is either wIse or Foolhardy.
∀x(((Cx ∧ Ox) ∧ Gx) → (Ix ∨ Fx))
Either all spotted giraffes are clever, or all clever giraffes are spotted.
∀x(Ox ∧ Gx → Cx) ∨ ∀x(Cx∧Gx → Ox)
Every clever giraffe is foolhardy.
∀x(Cx ∧ Gx → Fx)
If some giraffes are wise then not all giraffes are foolhardy.
∃x(Gx ∧ Ix) → ~∀x(Gx → Fx)
All giraffes are spotted if and only if no giraffes aren't spotted.
∀x(Gx → Ox) ↔ ~∃x(Gx ∧ ~Ox)
Nothing is both wise and foolhardy.
~∃x(Ix ∧ Fx)
SECTION 5c
1. a.
Only Friendly Elephants are Handsome (ambiguous)
i. ∀x(Hx → (Fx ∧ Ex))
ii. ∀x((Ex ∧ Hx) → Fx)
b. If only elephants are friendly, no Giraffes are friendly
∀x(Fx → Ex) → ~∃x(Gx ∧ Fx)
c. Only the Brave are fAir.
∀x(Ax → Bx)
d. If only elephants are friendly then every elephant is friendly
∀x(Fx → Ex) → ∀x(Ex → Fx)
e. All and only elephants are friendly.
All elephants are friendly [and] Only elephants are friendly.
∀x(Ex → Fx) ∧ ∀x(Fx → Ex)
f.
If every elephant is friendly, only friendly Animals are elephants (ambiguous)
i. ∀x(Ex → Fx) → ∀x(Ex → (Fx ∧ Ax))
ii. ∀x(Ex → Fx) → ∀x((Ex ∧ Ax) → Fx)
Copyrighted material
Chapter Three -- 58 -- Answers to the Exercises
Version of Aug 2013
Answers to Exercises -- Chapter 3
g.
h.
i.
j.
If any elephants are friendly, all and only giraffes are nasty
If some elephants are friendly, (all giraffes are Nasty and only giraffes are nasty)
∃x(Ex ∧ Fx) → (∀x(Gx → Nx) ∧ ∀x(Nx → Gx))
Among spOtted animals, only giraffes are handsome.
∀x(Ox → (Hx → Gx))
Among spotted animals, all and only giraffes are handsome
∀x(Ox → ((Gx → Hx) ∧ (Hx → Gx))
Only giraffes frolic if annoyed.
If a thing froLics if aNnoyed, it is a giraffe.
∀x((Nx → Lx) → Gx)
SECTION 5d
1. Symbolize these sentences.
a.
Every Giraffe which Frolics is Happy
∀x(Fx ∧ Gx → Hx)
b. Only giraffes which frolic are happy (ambiguous)
i. ∀x(Gx ∧ Hx → Fx)
ii. ∀x(Hx → Gx ∧ Fx)
c. Only giraffes are Animals which are Long-necked.
∀x(Ax ∧ Lx → Gx)
d. If only giraffes frolic, every animal which is not a giraffe doesn't frolic.
∀x(Fx → Gx) → ∀x(Ax ∧ ~Gx → ~Fx)
e. Some giraffe which frolics is long-necked or happy.
∃x((Fx ∧ Gx) ∧ (Lx ∨ Hx))
f.
No giraffe which is not happy frolics and is long-necked.
~∃x((~Hx ∧ Gx) ∧ (Fx ∧ Lx))
g. Some giraffe is not both long-necked and happy.
∃x(Gx ∧ ~(Lx ∧ Hx))
SECTION 5e
1. a.
b.
c.
d.
e.
f.
g.
h.
i.
If a Giraffe is Happy then it Frolics unless it is Lame.
∀x(Gx ∧ Hx → Fx ∨ Lx)
A Monkey frolics unless it is not happy.
∀x(Mx → Fx ∨ ~Hx)
Among giraffes, only happy ones frolic.
∀x(Gx → (Fx → Hx))
All and only giraffes are happy if they are not lame.
∀x(Gx ↔ (~Lx → Hx))
A giraffe frolics only if it is happy.
∀x(Gx ∧ Fx → Hx)
or
∀x(Gx → (Fx→Hx))
Only giraffes frolic if happy.
∀x((Hx → Fx) → Gx)
All monkeys are happy if some giraffe is.
∃x(Gx ∧ Hx) → ∀x(Mx → Hx)
Cute monkeys frolic.
∀x(Cx ∧ Mx → Fx)
Giraffes ruN and frolic if and only if they are Blissful and Exultant.
∀x(Gx → (Nx ∧ Fx ↔ Bx ∧ Ex))
Copyrighted material
Chapter Three -- 59 -- Answers to the Exercises
Version of Aug 2013
Answers to Exercises -- Chapter 3
j.
k.
l.
m.
n.
If those who are heAlthy are not lame, then if they are exultant, they will frolic.
∀x((Ax → ~Lx) → (Ex → Fx))
Only giraffes and monkeys are blissful and exultant.
∀x(Bx ∧ Ex → Gx ∨ Mx)
The brave(I) are happy.
∀x(Ix → Hx)
If a giraffe frolics, then no monkey is blissful unless it is.
∀x((Gx ∧ Fx) → (Bx ∨ ~∃y(My ∧ By)))
Giraffes and monkeys frolic if happy.
∀x(Gx ∨ Mx → (Hx → Fx))
SECTION 6
1. a.
The sky is Blue
Everything that is blue is prEtty
∴ Something is pretty
Be
∀x(Bx → Ex)
∴ ∃xEx
1
2
3
4
Show ∃xEx
Be → Ee
Ee
∃xEx
b.
Every Hyena is Grey.
Every hyena is an Animal
Jenny is a hyena
∴ Some animal is grey
pr2 ui
2 pr1 mp
3 eg dd
∀x(Hx → Gx)
∀x(Hx → Ax)
He
∴ ∃x(Ax ∧ Gx)
Copyrighted material
Chapter Three -- 60 -- Answers to the Exercises
Version of Aug 2013
Answers to Exercises -- Chapter 3
1
2
3
4
5
6
7
Show ∃x(Ax ∧ Gx)
He → Ge
He → Ae
Ge
Ae
Ae ∧ Ge
∃x(Ax ∧ Gx)
c.
If some Hyena is Grey, every hyena is grey
Every sCavenger is grey
Jenny is a hyena and a scavenger
Kathy is a hyena
∴ Kathy is grey
pr1 ui
pr2 ui
pr3 2 mp
pr3 3 mp
4 5 adj
6 eg dd
∃x(Hx ∧ Gx) → ∀x(Hx → Gx)
∀x(Cx → Gx)
He ∧ Ce
Ha
∴ Ga
1
2
3
4
5
6
7
8
9
10
11
Show Ga
Show ∃x(Hx ∧ Gx)
He
Ce
Ce → Ge
Ge
He ∧ Ge
∃x(Hx ∧ Gx)
∀x(Hx → Gx)
Ha → Ga
Ga
pr3 s
pr3 s
pr2 ui
4 5 mp
3 6 adj
7 eg dd
2 pr1 mp
9 ui
10 pr4 mp dd
2. The error is at line 3. It is not permissible to use EI to get an instance of pr2 in the variable z because z
occurs already on line 2; this would violate the restriction on EI.
3. No derivations are given for named theorems.
SECTION 8
1. Symbolize these arguments and provide derivations to validate them. Give an explicit scheme of
abbreviation for each.
a.
If history is right (P), then if anyone was strOng, hercules was strong.
Only those who work out (M) are strong, and only those with self-Discipline work out.
∴ If Hercules does not have self-discipline, then either history is not right or nobody is strong.
Copyrighted material
Chapter Three -- 61 -- Answers to the Exercises
Version of Aug 2013
Answers to Exercises -- Chapter 3
P → (∃xOx → Oh)
∀x(Ox → Mx) ∧ ∀x(Mx → Dx)
∴ ~Dh → (~P ∨ ~∃xOx)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Show ~Dh → (~P ∨ ~∃xOx)
~Dh
Mh → Dh
~Mh
Oh → Mh
~Oh
Show ~P ∨ ~∃xOx
Show ~~P → ~∃xOx
~~P
P
∃xOx → Oh
~∃xOx
~P ∨ ~∃xOx
b.
If some Giraffes are not Happy, then all giraffes are Morose.
Some giraffes pOnder the mysteries of life.
∴ If some giraffes are not morose, then some who ponder the mysteries of life are happy.
ass cd
pr2 s ui
2 3 mt
pr2 s ui
4 5 mt
ass cd
9 dn
p1 10 mp
6 11 mt cd
8 cdj dd
7 cd
∃x(Gx ∧ ~Hx) → ∀x(Gx → Mx)
∃x(Gx ∧ Ox)
∴ ∃x(Gx ∧ ~Mx) → ∃x(Ox ∧ Hx)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
Show ∃x(Gx ∧ ~Mx) → ∃x(Ox ∧ Hx)
∃x(Gx ∧ ~Mx)
Gi ∧ ~Mi
Show ~∀x(Gx → Mx)
∀x(Gx → Mx)
Gi → Mi
Mi
~Mi
~∃x(Gx ∧ ~Hx)
Gj ∧ Oj
Show Hj
~Hj
Gj ∧ ~Hj
∃x(Gx ∧ ~Hx)
~∃x(Gx ∧ ~Hx)
Oj ∧ Hj
∃x(Ox ∧ Hx)
Copyrighted material
ass cd
2 ei
ass id
5 ui
3 s 6 mp
3 s 7 id
4 pr1 mt
pr2 ei
ass id
10 s 12 adj
13 eg
9 r id
10 s 11 adj
16 eg cd
Chapter Three -- 62 -- Answers to the Exercises
Version of Aug 2013
Answers to Exercises -- Chapter 3
c.
There is not a single Critic who either Likes art or can pAint.
Some level-Headed peOple are critics.
Anyone who can't paint is unEducated.
∴ Some level-headed people are uneducated.
∀x(Cx → ~(Lx ∨ Ax))
∃x((Hx ∧ Ox) ∧ Cx)
∀x(Ox → (~Ax → ~Ex))
∴ ∃x((Hx ∧ Ox) ∧ ~Ex)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Show ∃x((Hx ∧ Ox) ∧ ~Ex)
(Hi ∧ Oi) ∧ Ci
Ci
Hi
Oi
Ci → ~(Li ∨ Ai)
~(Li ∨ Ai)
~Li ∧ ~Ai
~Ai
Oi → (~Ai → ~Ei)
~Ai → ~Ei
~Ei
(Hi ∧ Oi) ∧ ~Ei
∃x((Hx ∧ Ox) ∧ ~Ex)
d.
No Astronaut is a good Dancer.
Every sInger is warm-Blooded.
If something is warm-blooded and is not a good dancer, then nothing that is either a singer or an
astronaut is Exultant.
∴ If some astronaut is a singer, then no singer is exultant.
pr2 ei
2s
2ss
2ss
pr1 ui
3 6 mp
7 dm
8s
pr3 ui
5 10 mp
9 11 mp
2 s 12 adj
13 eg dd
∀x(Ax → ~Dx)
∀x(Ix → Bx)
∃x(Bx ∧ ~Dx) → ∀x((Ix ∨ Ax) → ~Ex)
∴ ∃x(Ax ∧ Ix) → ∀x(Ix → ~Ex)
Copyrighted material
Chapter Three -- 63 -- Answers to the Exercises
Version of Aug 2013
Answers to Exercises -- Chapter 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
Show ∃x(Ax ∧ Ix) → ∀x(Ix → ~Ex)
∃x(Ax ∧ Ix)
Show ∀x(Ix → ~Ex)
Show Ix → ~Ex
Ix
Ai ∧ Ii
Ai → ~Di
Ii → Bi
~Di
Bi ∧ ~Di
∃x(Bx ∧ ~Dx)
∀x((Ix ∨ Ax) → ~Ex)
(Ix ∨ Ax) → ~Ex
Ix ∨ Ax
~Ex
e.
All stuDents who have a sense of Humor or are Brilliant seek Fame.
Anyone who seeks fame and is brilliant is Insecure.
Whoever is a Mathematician is brilliant.
∴ Every student who is a mathematician is insecure.
ass cd
ass cd
2 ei
pr1 ui
pr2 ui
6 s 7 mp
6s 8 mp 9 adj
10 eg
11 pr3 mp
12 ui
5 add
13 14 mp cd
4 ud
3 cd
∀x((Dx ∧ (Hx ∨ Bx)) → Fx)
∀x(Fx ∧ Bx → Ix)
∀x(Mx → Bx)
∴ ∀x((Dx ∧ Mx) → Ix)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Show ∀x((Dx ∧ Mx) → Ix)
Show (Dx ∧ Mx) → Ix
Dx ∧ Mx
Dx
Mx
Mx → Bx
Bx
Hx ∨ Bx
Dx ∧ (Hx ∨ Bx)
(Dx ∧ (Hx ∨ Bx)) → Fx
Fx
Fx ∧ Bx
Fx ∧ Bx → Ix
Ix
Copyrighted material
ass cd
3s
3s
pr3 ui
5 6 mp
7 add
4 8 adj
pr1 ui
9 10 mp
7 11 adj
pr2 ui
12 13 mp cd
2 ud
Chapter Three -- 64 -- Answers to the Exercises
Version of Aug 2013
Answers to Exercises -- Chapter 3
f.
There is a Monkey that is Happy if and only if some Giraffe is happy.
There is a monkey that is happy if and only if some giraffe is not happy.
All monkeys are happy.
∴ It is not the case that either every giraffe is happy or none are.
∃x(Mx ∧ (Hx ↔ ∃x(Gx ∧ Hx))
∃x(Mx ∧ (Hx ↔ ∃x(Gx ∧ ~Hx))
∀x(Mx → Hx)
∴ ~(∀x(Gx → Hx) ∨ ∀x(Gx → ~Hx))
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
Show ~(∀x(Gx → Hx) ∨ ∀x(Gx → ~Hx))
∀x(Gx → Hx) ∨ ∀x(Gx → ~Hx)
Mi ∧ (Hi ↔ ∃x(Gx ∧ Hx)
Mj ∧ (Hj ↔ ∃x(Gx ∧ ~Hx)
Mi → Hi
Mj → Hj
Hi
Hj
∃x(Gx ∧ Hx)
∃x(Gx ∧ ~Hx)
Gk ∧ Hk
Gm ∧ ~Hm
Show ~∀x(Gx → Hx)
∀x(Gx → Hx)
Gm → Hm
Hm
~Hm
∀x(Gx → ~Hx)
Gk → ~Hk
~Hk
Hk
g.
For every Astronaut that writes pOetry, there is one that doesn't.
For every astronaut that doesn't write poetry, there is one that does.
∴ If there are any astronauts, some write poetry and some don't.
ass id
pr2 ei
pr3 ei
pr4 ui
pr4 ui
3 s 5 mp
3 s 6 mp
3 s bc 7 mp
4 s bc 7 mp
9 ei
10 ei
ass id
14 ui
12 s 15 mp
12 s id
2 13 mtp
18 ui
11 s 19 mp
11 s id
∀x((Ax ∧ Ox) → ∃x(Ax ∧ ~Ox))
∀x((Ax ∧ ~Ox) → ∃x(Ax ∧ Ox))
∴ ∃xAx → ∃x(Ax ∧ Ox) ∧ ∃x(Ax ∧ ~Ox)
Copyrighted material
Chapter Three -- 65 -- Answers to the Exercises
Version of Aug 2013
Answers to Exercises -- Chapter 3
1
Show ∃xAx → ∃x(Ax ∧ Ox) ∧ ∃x(Ax ∧ ~Ox)
2
ass cd
∃xAx
3
Ai
2 ei
4
T59
Oi ∨ ~Oi
5
Show Oi → ∃x(Ax ∧ Ox) ∧ ∃x(Ax ∧ ~Ox)
6
Oi
ass cd
7
3 6 adj
Ai ∧ Oi
8
7 eg
∃x(Ax ∧ Ox)
9
Pr1 ui
Ai ∧ Oi → ∃x(Ax ∧ ~Ox)
10
7 9 mp
∃x(Ax ∧ ~Ox)
11
8 10 adj cd
(∃x(Ax ∧ Ox) ∧ ∃x(Ax ∧ ~Ox))
12
Show ~Oi → ∃x(Ax ∧ Ox) ∧ ∃x(Ax ∧ ~Ox)
13
~Oi
ass cd
14
13 3 adj
Ai ∧ ~Oi
15
14 eg
∃x(Ax ∧ ~Ox)
16
Pr2 ui
Ai ∧ ~Oi → ∃x(Ax ∧ Ox)
17
14 16 mp
∃x(Ax ∧ Ox)
18
15 17 adj cd
(∃x(Ax ∧ Ox) ∧ ∃x(Ax ∧ ~Ox))
19
4 5 12 sc
∃x(Ax ∧ Ox) ∧ ∃x(Ax ∧ ~Ox)
20
19 cd
<Could also skip line 4 and use sc appealing only to lines 5 and 12.>
SECTION 9
1. a.
~∃x(Ax ∨ Bx)
∀x∀y(Gx ∧ Hy → By)
∃xGx
∴ ∀x~Hx
1
2
3
4
5
6
7
8
9
10
11
12
Show ∀x~Hx
~∀x~Hx
∃xHx
Hi
Gj
Gj ∧ Hi
Gj ∧ Hi → Bi
Bi
∀x~(Ax ∨ Bx)
~(Ai ∨ Bi)
~Ai ∧ ~Bi
~Bi
Copyrighted material
ass id
2 qn
3 ei
pr3 ei
4 5 adj
pr2 ui ui
6 7 mp
pr1 qn
9 ui
10 dm
11 s 8 id
Chapter Three -- 66 -- Answers to the Exercises
Version of Aug 2013
Answers to Exercises -- Chapter 3
b.
∃x(Hx ∧ ~∃y(Gy ∧ Hx))
∴ ∀y~Gy
1
2
3
4
5
6
7
8
9
Show ∀y~Gy
~∀y~Gy
∃yGy
Hi ∧ ~∃y(Gy ∧ Hi)
Hi
~∃y(Gy ∧ Hi)
Gj
Gj ∧ Hi
∃y(Gy ∧ Hi)
c.
∀x(Ax → ∀y(Bx ↔ By))
∃zBz
∴ ∀y(Ay → By)
1
2
3
4
5
6
7
8
9
10
11
Show ∀y(Ay → By)
Show ∀i(Ai → Bi)
Show Ai → Bi
Ai
Ai → ∀y(Bi ↔ By)
∀y(Bi ↔ By)
Bj
Bi ↔ Bj
Bi
d.
~∀x(Dx ∨ Ex)
∃x(Fx ↔ ~Ex) → ∀zDz
∴ ∃x~Fx
ass id
2 qn
pr1 ei
4s
4s
3 ei
5 7 adj
8 eg 6 id
∀y(Ay → By)
Copyrighted material
ass cd
pr1 ui
4 5 mp
pr2 ei
6 ui
8 bc 7 mp cd
3 ud
2 av dd
Chapter Three -- 67 -- Answers to the Exercises
Version of Aug 2013
Answers to Exercises -- Chapter 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Show ∃x~Fx
~∃x~Fx
∀xFx
∃x~(Dx ∨ Ex)
~(Di ∨ Ei)
~Di ∧ ~Ei
~Di
~Ei
Show Fi → ~Ei
~Ei
Show ~Ei → Fi
Fi
Fi ↔ ~Ei
∃x(Fx ↔ ~Ex)
∀zDz
Di
e.
Jc ∧ ~Jd
∀xKx ∨ ∀x~Kx
∃x(Jx ∧ Kx) → ∀x(Kx → Jx)
∴ ~Kc
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Show ~Kc
Kc
Show ~∀x~Kx
∀x~Kx
~Kc
Kc
∀xKx
Jc ∧ Kc
∃x(Jx ∧ Kx)
∀x(Kx → Jx)
Kd → Jd
Kd
Jd
~Jd
Copyrighted material
ass id
2 qn
pr1 qn
4 ei
5 dm
6s
6s
8 r cd
3 ui cd
9 11 cb
13 eg
14 pr2 mp
15 ui 7 id
ass id
ass id
4 ui
2 r 5 id
3 pr2 mtp
pr1 s 2 adj
8 eg
9 pr3 mp
10 ui
7 ui
11 12 mp
pr1 s 13 id
Chapter Three -- 68 -- Answers to the Exercises
Version of Aug 2013
Answers to Exercises -- Chapter 3
SECTION 10
1. a.
∀x(Ax → ∃y(By ∧ ~Ay))
~∀xBx
~∃x(Bx ∧ Cx)
∴ ∃x(Ax ∧ Cx)
Universe: {1, 2, 3}
A: {1}
B: {2}
C: {3}
b.
∃x(Dx ∧ Ex ∧ ~Fx)
∃x(~Dx ∧ ~Ex)
∀x(Ex → Dx ∨ Fx)
∴ ∀x(Dx ∧ Ex → ~Fx)
Universe: {1, 2, 3}
D: {1, 2}
E: {1, 2}
F: {1}
c.
∃x(Fx ∧ Gx)
∃x(Fx ∧ ~Gx)
∃x(~Fx ∧ Gx)
∴ ∀x(~Fx → Gx)
<requires more than three things in the universe>
Universe: {1, 2, 3, 4}
F: {2, 3}
G: {1, 2}
d.
∀x∃y(Fx ↔ (Gy ∨ Fx))
∴ ~∃xFx → ~∃xGx
Universe: {1, 2}
F: { }
G: {1}
e.
Ha ∧ ~Hb
∀x(Kx → Hx ∧ Jx)
∃x(Jx ∧ ~Kx)
∴ ∃x(Hx ∧ ~Jx)
Universe: {1, 2}
H: {1}
J: {1,2}
K: { }
a --- 1
b --- 2
Copyrighted material
Chapter Three -- 69 -- Answers to the Exercises
Version of Aug 2013
Answers to Exercises -- Chapter 3
SECTION 11
1. For each of the following arguments use the method of expansions to determine whether the following
is a counterexample for it or not.
Universe:
0
1
2
F: {0}
G: {0, 2}
H: {2}
a: 2
b: 0
a.
∀x(Hx → ∃y(Fy ∧ ~Hy))
~∀xFx
~∃x(Fx ∧ Gx)
∴ ∃x(Hx ∧ Gx)
The conclusion expands to:
(Ha1 ∧ Ga1) ∨ (Ha2 ∧ Ga2) ∨ (Ha3 ∧ Ga3)
which is true because Ha3 and Ga3 are true. Since we have a true conclusion, we don't have a
counterexample.
b.
∃x(Gx ∧ Hx ∧ ~Fx)
∃x(~Gx ∧ ~Hx)
∀x(Hx → Gx ∨ Fx)
∴ ∀x(Gx ∧ Hx → ~Fx)
The conclusion expands to:
(Ga1 ∧ Ha1 → ~Fa1) ∧ (Ga2 ∧ Ha2 → ~Fa2) ∧ (Ga3 ∧ Ha3 → ~Fa3)
which is true because the first conjunct has a false antecedent, the second conjunct has a false
antecedent, and the third conjunct has a true consequent. Since we have a true conclusion, we don't
have a counterexample.
c.
∃x(Fx ∧ Gx)
∃x(Fx ∧ ~Gx)
∃x(~Fx ∧ Gx)
∴ ∀x(~Fx → Gx)
The second premise expands to:
(Fa1 ∧ ~Ga1) ∨ (Fa2 ∧ ~Ga2) ∨ (Fa3 ∧ ~Ga3)
which is false because each disjunct is false. Since we have a false premise we don't have a
counterexample.
Copyrighted material
Chapter Three -- 70 -- Answers to the Exercises
Version of Aug 2013
Answers to Exercises -- Chapter 3
d.
∀x∃y(Fx ↔ (Gy ∨ Fx))
∴ ~∃xFx → ~∃xGx
The conclusion expands to:
~(Fa1 ∨ Fa2 ∨ Fa3) → ~(Ga1 ∨ Ga2 ∨ Ga3)
which is true because the antecedent is false because its leftmost disjunct is true. Since we have a true
conclusion we don't have a counterexample.
e.
Ha ∧ ~Hb
∀x(Fx → Hx ∧ Gx)
∃x(Gx ∧ ~Fx)
∴ ∃x(Hx ∧ ~Gx)
The second premise expands to:
(Fa1 → Ha1 ∧ Ga1) ∧ (Fa2 → Ha2 ∧ Ga2) ∧ (Fa3 → Ha3 ∧ Ga3)
which is false because the first conjunct has a true antecedent and a false consequent. Since we have a
false premise, we don't have a counterexample.
Copyrighted material
Chapter Three -- 71 -- Answers to the Exercises
Version of Aug 2013