The Truth Table Method & The Truth Tree Method
1 Introduction
1. Unlike the Big 8 Method and like the Method of Derivation, the Truth Table
method and the Truth Tree method can be applied to inferences of any length; each will
work no matter how many (or how few) premises an inference has.
2. The method of Derivation tells us that, if we find a derivation, the conclusion
follows validly from the premises, but it remains silent when we cannot find a derivation;
perhaps we have failed to find a derivation because there indeed is no derivation, or
because of lack of skill on our part. The Truth Table method and the Truth Tree method
give us ways of determining that an inference is valid, if it is valid, or invalid if it is invalid.
3. In addition, the truth table and truth tree methods are purely mechanical
methods, which do not rely on any ingenuity on our part. However, the truth table method
is artificial in that it does not demonstrate the steps in reasoning by which the conclusion
is reached. The truth tree method solves this problem to some extent.
4. Another disadvantage of the truth table method, but less so of the truth tree
method, is that truth tables can also be quite cumbersome to produce. A shorter version
of the truth table method, called the targeted truth table method, is also given.
5. Both the truth table and truth tree methods are used to evaluate inferences
whose propositions are in symbolic language.
2 Truth Values & Truth Tables For The Logical Operators
1. A simple proposition can take one of two truth values, true (T) or false (F). (It's
a little strange to say, but 'false' is a 'truth value'.) Whether a well-formed complex
proposition is true or false depends solely upon whether the simple propositions in it are
true or false, according to the following four tables:
~ S
S v T
S & T
S ⊃ T
F T
T F
T
T
F
F
T
T
F
F
T
T
F
F
T
T
T
F
T
F
T
F
T
F
F
F
T
F
T
F
T
F
T
T
T
F
T
F
The tables are read like this:
Tilde: If "S" (i.e. any well-formed proposition) is true — represented by the nonbold letter "T" directly under the "S" — then "~S" is false — the bold "F" directly under the
tilde. And, in the second row, if "S" is false, then "~S" is true.
Wedge: If "S" is true and "T" is true — the non-bold "T"s directly under "S" and "T"
— then "S v T" is true — the bold "T" directly under the wedge. In the second row: If "S"
is true and "T" is false then "S v T" is true. In the third row: if "S" is false and "T" is true
then "S v T" is true. In the fourth: if "S" is false and "T" is false then "S v T" is false.
Ampersand: First row: if "S" is true and "T" is true then "S & T" is true. Second row:
if "S" is true and "T" is false then "S & T" is false. Third row: if "S" is false and "T" is true
then "S & T" is false. Fourth: if "S" is false and "T" is false then "S & T" is false.
Horseshoe: First row: if "S" is true and "T" is true then "S ⊃ T" is true. Second row:
if "S" is true and "T" is false then "S ⊃ T" is false. Third row: if "S" is false and "T" is true
then "S ⊃ T" is true. Fourth: if "S" is false and "T" is false then "S ⊃ T" is true.
2. The truth table for tilde (negation) is obvious, given the assumption that there
are two and only two truth values, true and false. If the proposition "Salt is an ingredient
in ketchup." is true, then the proposition "Salt is not an ingredient in ketchup." is false.
And if the proposition "Salt is an ingredient in ketchup." is false, then the proposition
"Salt is not an ingredient in ketchup." is true.
The truth table for wedge (disjunction) indicates that a proposition is false when
both disjuncts are false (in the fourth row); otherwise (in the other three) it is true. This
truth table allows that a disjunction is true when both disjuncts are true. For example, the
sign at a charity book sale "Everyone is welcome to donate goods or buy them." does not
prohibit anyone from doing both. This interpretation is called "inclusive-or". Propositions
which disallow the 'both' option, such as "you can have ice-cream or chocolate but not
both." or "Jack will take care of Jim or else Gill will.", are called "exclusive-or". These can
be translated into symbolic as "(S v T) & ~(S & T)".
The truth table for ampersand (conjunction) shows (first row) that a conjunction
is true only when both conjuncts are true; otherwise it is false. For example, "Jack is in
Baghdad and Gill is in Boston." is false if one or both of the conjuncts is false.
The truth table for horseshoe (conditional) shows that a conditional is false only
when the antecedent is true and the conditional is false (second row); otherwise it is true
(rows one, three, and four). Rows one and two are perhaps obvious. Consider the
following example: "If I take the full course of antibiotics, the infection will clear up.".
This complex proposition is true if I take the full course and the infection clears, and is
false if I take the full course of antibiotics and the infection persists. Rows three and four
require some explanation. These last two lines both have a false antecedent, and also state
that the conditional is true, regardless of the truth or falsity of the consequent. If I do not
take the full course of antibiotics, is the conditional false? No, but one might think that it
is not true, either. Why should we default to "T"?
"If …, then …" is often understood causally, as in the example of the antibiotic and
the infection. But "If …, then …" has other uses in addition to expressing causal
connection, such as to express logical connections ("If "a ⊃ b" and "a", then "b"."), or
connections based on definitions ("If it's water, it has hydrogen."). Conditionals are also
used to speculate about what might be, as in "If I were to win the Lotto, I would quit my
job.". (One can even use "If …, then …" to strongly assert the falsity of an antecedent, as
in "If United loses on Saturday, then I'll eat my hat.").
All of these uses have it in common that the conditional is false if the antecedent is
true and the consequent false. It might thus be better to think of "S ⊃ T" not as "If S, then
T" but as "not both S and not T", or, in symbolic, "~(S & ~T)". In other words, that the
conditional is false when the antecedent is true and the consequent is false is equivalent
to saying that it can't simultaneously be that the antecedent is true and the consequent is
false. When we think of the horseshoe in this way, we get the same results as in the truth
table for horseshoe, above. If "S" is true and "T" is false (i.e. the second row), "~T" is true
and "S & ~T" is true, and ~(S & ~T) is false. Under all other assignments (T and T, F and
T, F and F), the proposition is true. And note that when "S" is false (as it is in the bottom
of two lines of the truth table for the horseshoe), the internal conjunction is false, and the
negation is true. Thus, "S ⊃ T" is equivalent to "~(S & ~T)".
"S ⊃ T" and "~(S & ~T) are, further, equivalent to "~S v T", as in "Either I do not
take the full course of antibiotics or the infection will clear up.". Again, when "S" is true
and "T" is false, the proposition is false, and when "S" is false (as it is in the bottom two
lines of the truth table for the horseshoe), "~S" is true, and this is enough to make the
disjunction true.
Understood in this truth-functional way, the horseshoe is called material
conditional or material implication, since it is agnostic about the type of connection being
asserted in the original English proposition and merely expresses the implication between
two propositions.
3. Since the four types of complex proposition are composed of simple
propositions and the four logical operators, we can work out the truth value of a
proposition based on the truth values of the simple propositions and the four basic truth
tables, given above. Consider the following proposition:
((p ⊃ q) & r) v ~q
T
F
T
F
Suppose that the truth values of the simple propositions are as follows: "p" is true,
"q" is false, and "r" is true. What is the truth value of the proposition as a whole? The trick
to determining the truth value of a complex proposition is to work from the inside(s) out.
By "the insides" we mean most heavily nested operator(s), the operator(s) buried within
the greatest number of pairs of parentheses. In this proposition, the horseshoe has two
pairs of parentheses around it, the ampersand has one, the negation has one (this pair of
parentheses is invisible because of the rule that a tilde negates whatever follows it
immediately!), and the wedge has none. Thus we will begin with the horseshoe and then
move to the ampersand. At this stage, we will have a value for the left-hand disjunct. In
the right-hand disjunct, we must deal with the negation. Finally, we can work out the truth
value for the whole proposition by turning to the wedge, which is the main operator.
Beginning with the horseshoe and "p ⊃ q", since we are supposing for the sake of
example that "p" is true and "q" is false, "p ⊃ q" is false; the second row of the truth table
for horseshoe tells us this: propositions having the form "S ⊃ T" are false when "S" is true
and "T" is false. This value (F, in bold, below the horseshoe) now stands for the
proposition "p ⊃ q".
((p ⊃ q) & r) v ~q
T F F
T
F
Moving to the ampersand, "p ⊃ q" is the left-hand conjunct and its value is F, as
we have just worked out. Given that "r" is true, "(p ⊃ q) & r" is false, since propositions
having the form "S & T" are false when "S" is false and "T" is true. This value (F, in bold,
below the ampersand) now stands for the proposition "(p ⊃ q) & r".
((p ⊃ q) & r) v ~q
T F F F T
F
Moving to the disjunction, we see that the right-hand disjunct is a negation and we
must deal with the negation before the disjunction. Since the proposition being negated
"q") is false, it is true, for propositions having the form "~S" are true when "S" is false.
((p ⊃ q) & r) v ~q
T F F F T
TF
Finally, we put the values for the left-hand and right-hand disjuncts together in the
disjunction. "((p ⊃ q) & r) v ~q" is true, for propositions having the form "S v T" are true
when "S" is false and "T" is true.
((p ⊃ q) & r) v ~q
T F F F T T TF
This means that when "p" is true, "q" is false and "r" is true, the complex proposition
"((p ⊃ q) & r) v ~q" is true.
Exercises
For each proposition, underline "T" if it is true and "F" if it is false, when "b" is true, "c"
is false, "d" is true, "e" is false, and "f" is true.
b c d e f
T F T T T
Sample
d ⊃
T
T
T
T T
(b & ~c)
T
F
T
TF
T T TF
T T TF
T F
(1) ~~b
T F
(2) c ⊃ d
T F
(3) (e v ~f) ⊃ d
T F
(4) ~((e & ~f) ⊃ b)
T F
(5) f ⊃ (b ⊃ e)
T F
(6) ~(b ⊃ ~c) & (e ⊃ d)
T F
(7) ~f v (((b ⊃ c) ⊃ (~f & ~e)) v ~~d)
T F
(8) (~b v (b ⊃ c)) ⊃ ((~f & ~e) v ~d)
T F
Answers to Even Numbers
For each proposition, underline "T" if it is true and "F" if it is false, when "b" is true, "c" is
false, "d" is true, "e" is false, and "f" is true.
(2) c ⊃ d
c ⊃ d
F T T
T
(4) ~((e & ~f) ⊃ b)
~((e
F
F
F
F
F F
F
& ~f) ⊃ b)
T
T
FT
T
F FT
T
F FT T T
F FT T T
(6) ~(b ⊃ ~c) & (e ⊃ d)
~(b
T
T
T
F T
F T
F T
⊃ ~c) & (e ⊃ d)
F
F
T
TF
F
T
T TF
F
T
T TF
F
T
T TF
F T T
T TF F F T T
F
(8) (~b v (b ⊃ c)) ⊃ ((~f & ~e) v ~d)
(~b
T
FT
FT
FT
FT
FT
FT
v (b
T
T
T
F T
F T
F T
F T
⊃ c)) ⊃ ((~f & ~e) v ~d)
F
T
F
T
F
FT
TF
FT
F F
FT
TF
FT
F F
FT
TF
FT
F F
FT F TF
FT
F F
FT F TF F FT
F F
T
FT F TF F FT
<----
T
(In fact, we can stop after the fourth line: once the antecedent is F, the conditional must be
T)
3 Setting Up Truth Tables
1. In the previous section, we gave each of the simple propositions a truth value
and used these to work out the value of a complex proposition. When we use The Truth
Table Method, we do not know the values of the simple propositions and so imagine all of
the possible combinations of values (also called assignments) and we work out the value
of a complex proposition for each assignment. Moreover, we have to do this for all of the
propositions in an inference. Luckily, there is a simple way to do both of these at once,
called a truth table.
2. In this section, we present a procedure for setting up a truth table which will
ensure that (a) no combination of truth values is duplicated and (b) no possible
combination is left out.
First, determine how many rows the table will have. You can do this once you have
put the passage into standard form using symbolic. Count up how many distinct simple
propositions there are in the passage. Even if a letter appears more than once, count it
only once. The following for example, involves only three distinct letters, even though "p"
and "r" each occur twice.
1. (p v q) ⊃ r
2. p
-3. r
If the propositions of an inference involve only 1 distinct letter, then the truth table
has 2 rows. If they involve 2 distinct letters in it, then the truth table has 4 rows. If 3
letters, 8 rows. If 4 letters, 16 rows. And so on. In general, if an inference involves n letters,
then its truth table has 2n rows. Suppose that the propositions of an inference involve 3
letters:
p
q
r
Given that it has 3 letters, there will be 8 rows in the truth table, as 23 = 8.
Second, fill in each column, as follows: We begin with the first simple proposition
and under it write down (n/2 = 8/2 =) 4 Ts and then 4 Fs.
The table at this stage looks like this:
p
q
r
T
T
T
T
F
F
F
F
For the next column, we divide by 2 again to get (4/2 =) 2 and so, below the second
proposition letter we alternate between pairs of Ts and Fs.
p
q
T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
r
Dividing by 2 again to get (2/2 =) 1, under the third letter we alternate between single Ts
and Fs. The complete assignment appears as follows:
p
q
r
T
T
T
T
T
F
T
F
T
T
F
F
F
F
F
T
T
F
F
F
T
F
T
F
In general terms, the procedure runs as follows. First, determine the number of
rows in the truth table. Let that number be n. Second, give a column of n/2 Ts under the
first letter, and follow this with an equal number of Fs. At this point, the first column is
done. Third, give a column of Ts under the second letter half as long as the column of Ts
under the first, give a column of Fs under the Ts equal in length, give a column of Ts under
the Fs equal in length, and give a column of Fs under the Ts equal in length. At this point,
the second column is done. And so on.
3. Once we have made sure that we will be considering every possible assignment
of truth and falsity to the simple propositions involved, we can work out what the value
of each whole proposition is under each assignment.
4 The Truth Table Method
1. The truth table method works like this: First, convert the inference into symbolic
and put it in standard form. Second, make a truth table for the inference. Third, check the
table for rows in which all of the premises are true and the conclusion is false. If there is
such a row (or rows) the inference is invalid; if not, it is valid.
The truth table method rests on the fact that an inference is valid if it is impossible
for the premises to be true and the conclusion false. If the truth table shows that it is
possible for the premises to be true and the conclusion false, the inference is not valid. If
this combination of true premises and false conclusion is impossible, the inference is
valid.
2. Consider the following inference:
Jones owns a Ford. So, either Jones owns a Ford, or Brown is in Barcelona.
With "o" standing for the proposition "Jones owns a Ford.", with "b" standing for "Brown
is in Barcelona." in standard form, using symbolic, the inference is written as follows:
1. o
-2. o v b
In the truth table method, we make a truth table, as described above, for all of the
propositions in the inference at once. We begin by picking out the simple propositions.
This inference involves two simple propositions, "o" and "b". On one line, we first write
down the simple propositions and then the premises and the conclusion. In this case,
there is one premise and the conclusion. Thirdly, we generate all of the possible truth
assignments to the simple propositions. Since there are two simple propositions, the table
has four rows. The final step is to work out the truth values of the propositions of the
inference, according to the basic rules and the procedure above. The final truth table looks
like this:
o
b
o
T
T
F
F
T
F
T
F
T
T
F
F
o
b
v
T
T
T
F
(If you find it helpful, you can fill in the values for each proposition everywhere it
appears. In this inference, you could fill them in under "o" and "b" in "o v b". The final
truth table would look like this rather than the one just above:
o
b
o
o
v
b
T
T
F
F
T
F
T
F
T
T
F
F
T
T
F
F
T
T
T
F
T
F
T
F
)
We put the column of values under each of the propositions in bold (or when
writing by hand, draw a box around it). For complex propositions, this column should be
under the main operator. In this inference, it appears under the wedge in the conclusion.
Finally, we inspect the table in search of a row in which the premise is true and the
conclusion is false. If we find such a row, the inference is invalid, since we know from our
earlier discussion of the concept of 'validity' that no inference with true premises and a
false conclusion can be valid. The premise is true in the first and second rows, and so is
the conclusion. In the third and fourth rows, the premise is false. So there is no row in
which the premise is true and the conclusion is false and, thus, this inference is valid.
2. Here is a truth table for an instance of MA, the inference "p ⊃ q. p. So, q.". The
first step is to identify the simple propositions. In this case, there are two: "p" and "q". We
write these down in a row. The next (second) step is to write down the propositions of the
inference, on the same line as the simple propositions. After the second step, the truth
table looks like this:
p
q
(p
⊃
q)
p
q
The third to step is to generate the possible combinations of assignments of T and F to the
simple propositions. Since there are two simple propositions, there are four possible
assignments of truth values. After step three, the truth table looks like this:
p
q
T
T
F
F
T
F
T
F
(p
⊃
q)
p
q
The fourth step is to work out the truth values of the propositions for each assignment, as
described above. After this fourth step, the table looks like this:
p
q
T
T
F
F
T
F
T
F
(p
q)
⊃
T
F
T
T
p
q
T
T
F
F
T
F
T
F
There is no row in which the premises are true and the conclusion false. Thus, this
inference is valid.
3. In contrast to the two examples so far, this inference is invalid:
1. b
-2. b & c
Its truth table looks like this:
b
c
b
T
T
F
F
T
F
T
F
T
T
F
F
b
&
c
T
F
F
F
<----
In the second row, the premise is true while the conclusion is false. We point to any
invalidating row with an arrow.
4. Now consider the following inference:
If Henry was not required to work and was able to save up enough money, he will
be at the concert tonight. He's not at the concert. So, he must not have been able
to save up the money.
With "w" standing for "Henry was required to work.", "s" standing for "Henry was able to
save up enough money.", with "c" standing for "Henry is at the concert." in standard form,
using symbolic, it looks like this:
1. (~w & s) ⊃ c
2. ~c
--------------3. ~s
This inference involves three distinct propositions, and so its truth table has eight rows.
The completed table looks thus:
w
s
c
(~w
&
T
T
T
F
F
s)
⊃
T
c
~c
~s
F
F
T
T
T
F
F
F
F
T
F
F
T
T
F
F
F
T
F
T
F
T
F
F
F
F
T
T
T
T
F
F
F
T
T
F
F
T
T
T
T
F
T
T
T
F
T
F
T
F
T
F
T
T
F
F
T
T
<----
In the second row, the premises are true while the conclusion is false. The inference, thus,
is invalid.
Exercises
Sample
Use the truth table method to show that some arbitrary instance of CC is valid.
1. a ⊃ b
a b a ⊃ b ~b ~a
2. ~b
T T
T
F F
-----T F
F
T F
3. ~a
F T
T
F T
F F
T
T T
No line has true premises and a false conclusion. So, the inference is valid.
(1)
(2)
(3)
(4)
(5)
(6)
Use the truth table method to show that some arbitrary instance of Chain is
valid.
Use the truth table method to show that some arbitrary instance of MC is not
valid.
Use the truth table method to show that some arbitrary instance of CA is not
valid.
Use the truth table method to show that some arbitrary instance of Conj. is
valid.
Use the truth table method to show that some arbitrary instance of CD is valid.
Use the truth table method to determine whether the following inference is
valid or not valid.
1. (b v ~c) ⊃ d
2. ~c
------------3. d
(7)
Consider the following inference:
If national elections deteriorate into television popularity contests, smooth-talking
morons will get elected. So clearly, smooth-talking morons won't get elected if the
elections don't deteriorate into television popularity contests.
(a) Make a translation key for it.
(b) Relative to the key, translate the inference into symbolic and put it in standard form.
(c) Relative to what you have in standard form, make a truth table for the inference.
(d) Relative to your truth table, is the inference valid or not valid?
Answers to Even Numbers
(2)
Use the truth table method to show that some arbitrary instance of MC is not valid.
1. a ⊃ b
2. b
-----3. a
a
T
T
F
F
b
T
F
T
F
a ⊃ b
T
F
T
T
b
T
F
T
F
a
T
T
F <---F
(4) Use the truth table method to show that some arbitrary instance of Conj. is valid.
1. a
2. b
-3. a & b
a
T
T
F
F
b
T
F
T
F
a
T
T
F
F
b
T
F
T
F
a & b
T
F
F
F
(6) Use the truth table method to determine whether the following inference is valid or not
valid.
1. (b v ~c) ⊃ d
2. ~c
------------3. d
b
T
c
T
d
T
(b v ~c)⊃ d
T F T
~c
F
d
T
T
T
T
F
F
F
F
T
F
F
T
T
F
F
F
T
F
T
F
T
F
T
T
T
F
F
T
T
F
T
T
F
F
T
T
F
T
F
T
T
T
F
F
T
T
F
F
T
T
F
T
F
T
F
T
F
Valid – there is no line with T premises (the horseshoe in the first proposition and the
tilde in the second) and F conclusion
5 Logical Equivalence & Inequivalence, & Logical Contradiction
1. The proposition "If Jones is in Columbus, then he is in Ohio." is logically
equivalent to "If Jones is not in Ohio, then he is not in Columbus.". Truth tables bear this
out—the two tables have the same column of final values. With "c" standing for "Jones is
in Columbus." and with "o" standing for "Jones is in Ohio.", the two become:
c ⊃ o
~o ⊃ ~c
A truth table for both propositions simultaneously looks like this:
c
o
T
T
F
F
T
F
T
F
c
⊃
o
⊃
~c
F
T
F
T
T
F
T
T
F
F
T
T
T
F
T
T
~o
There is no row in which the two propositions have different truth values. So, they are
logically equivalent.
2. In contrast, b ⊃ c and c ⊃ b are logically inequivalent. Consider the following
table:
b
c
T
T
F
F
T
F
T
F
b
⊃
T
F
T
T
c
c
⊃
b
T
T
F
T
Not every row has the same ultimate value; rows 2 and 3 both have opposite truth values.
3. If two propositions have opposite values on every line, they are said to be logical
contradictories.
Exercises
(1)
Make a truth table for the propositions "p" and "~~p" and, relative to your truth
table, determine whether they are logically equivalent or logically inequivalent.
(2) Make a truth table for the propositions "~(b & c)" and "~b v ~c", and, relative to your
truth table, determine whether they are logically equivalent or logically
inequivalent.
(3) Make a truth table for the propositions "~(b & c)" and "~b & ~c" and, relative to your
truth table, determine whether they are logically equivalent or logically
inequivalent.
(4)
Make a truth table for the propositions "b ⊃ c" and "~b v c", and, relative to your
truth table, determine whether they are logically equivalent or logically
inequivalent.
(5) Make a truth table for the propositions "p ⊃ (q ⊃ r)" and "(p & q) ⊃ r" and, relative
to your truth table, determine whether they are logically equivalent or logically
inequivalent.
(6) Make truth tables for the sentences "a & (b v c)" and "(a & b) v (a & c)" and, relative
to your truth table, determine whether they are logically equivalent or logically
inequivalent.
6 Targeted Truth Tables
1. Truth tables can be cumbersome, especially when there are multiple simple
propositions. One way to shorten the process is by using the targeted truth table method.
This method relies on the fact that in the truth table method we are interested in lines of
the truth table in which the premises are true and the conclusion is false. The targeted
truth table method attempts to find an assignment (or assignments) on which the
premises are true and the conclusion is false, without making a full truth table. If such an
assignment(s) can be found, the inference is invalid; if not, the inference is valid.
2. Consider the following inference, in standard form using symbolic:
1. b ⊃ c
2. a & (c v d)
-----------3. d v b
Since there are four distinct simple propositions, the full truth table would have 16
lines. We can shorten the process by using the targeted truth table method. We begin by
writing the simple propositions, the premises and conclusion on a line, as follows:
a b c d
b ⊃ c
a
&
(c v d)
d v b
Since we are targeting the row(s) of the table on which the conclusion is false and
the premises true, we think about which assignment(s) would make the conclusion false,
or any of the premises true. In this example, the conclusion makes a good place to begin,
since a disjunction is false only when both disjuncts are false. So we assign F to both "d"
and "b" in the conclusion, and everywhere else either "d" or "b" appears in the premises.
We fill in these values for "b" and "d" wherever they appear and we write an "F" under the
wedge in the conclusion. At this stage, the targeted truth table looks like this:
a b c d
F
F
b ⊃ c
F
a
&
(c v d)
F
d v b
F F F
Using this assignment for "d" and "b", can an assignment be found for the
remaining letters ("a" and "c") such that the premises are true? The first premise is a
conditional, with a false antecedent, so it will be true regardless of the value we give to
"c". We can thus leave the first premise, and the value of "c", aside for the time being. The
second premise is a conjunction. In order for it to be true both conjuncts must be true. So
we assign T to "a" and T to "c".
a b c d
T F T F
b ⊃ c
F T T
a
T
&
T
(c v d)
T T F
d v b
F F F
We have thus shown that the inference is invalid, because when "a" is T, "b" is F,
"c" is T and "d" is F, the premises are true and the conclusion false.
3. Consider the following, different, inference:
1. b & c
2. a ⊃ (c v d)
-----------3. d v b
Employing the same procedure as before, we write out the propositions involved,
and assign F to "d" and "b" in order to make the conclusion F. This time however, we can
see not only that the conclusion is F (we write an "F" under the wedge in the conclusion)
but also that the first premise will be F (even without knowing the value of "c". So, we can
write an "F" under the ampersand.
a b c d
F
F
b & c
F F
a ⊃ (c v d)
F
d v b
F F F
Since the first premise cannot be true when we make the conclusion false (F) by
assigning F to both b and d, (because for a conjunction to be true, both conjuncts must be
true) this inference is valid; it is impossible for the premises to be true and the conclusion
false.
4. In general, we start by looking for assignments that we must make in order to
make either a premise true or the conclusion false; these will give us fixed pieces of
information. We can immediately assign values to any propositions that are not complex.
If a premise is a single proposition letter, assign T to it; if a conclusion is a single
proposition letter, assign F to it. After this, it is a good idea to begin with the conclusion,
since there is only ever one conclusion and it must be made false. Moreover, propositions
which are either negations, disjunctions or conditionals are false under only one
assignment. A negation is false when what is negated is true; a disjunction is false when
both disjuncts are false; and, a conditional is false only when the antecedent is true and
the consequent is false.
5. However, if the conclusion is a complex statement using parentheses, the best
strategy might not be to begin with the conclusion. For example, if the conclusion is "~(a
⊃ b)", we know that this is false when "a ⊃ b" is true. But there are multiple assignments
under which "a ⊃ b" is true. Conjunctions, similarly, are false when either, or both, of the
conjuncts are false.
In these cases, we should look to the premises. For example, consider the following
inference:
1. b & c
2. b v d
3. d ⊃ a
------4. ~(a ⊃ b)
When we write out the propositions on a line, we see that there is no single
assignment under which the conclusion is false. When we look at the premises, we are
confronted with a conjunction, a disjunction and a conditional. Conjunctions, which are
false under multiple assignments, are true only when both conjuncts are true, so the
premise to begin with is the first one: we assign T to both "b" and "c" and fill in what else
we can:
a b c d
T T
b & c
T T T
b v d
T T
d ⊃ a
~(a ⊃ b)
F
T T
The second premise is already true, so we leave the value of "d" in the second
premise aside for the moment. We are not forced into making an assignment to "d" based
on the third premise either, since if "a" were true, it wouldn't matter what the value of "d"
was. What about "a"? Again, we are not forced into a value for "a". Based on the third
premise, it could be either, since "d" could be either; based on the conclusion, it could be
either, since "b" is T. So there are a number of assignments which would show this
inference to be invalid. If we assign T to "a", "d" can be either T or F, as follows:
a b c d
T T T
b & c
T T T
b v d
T T
d ⊃ a
T T
~(a ⊃ b)
F T T T
Alternatively, if we assign F to "d", "a" can be either T or F, as follows:
a b c d
T T F
b & c
T T T
b v d
T T F
d ⊃ a
F T
~(a ⊃ b)
F
T T
Any one of the following assignments of values is sufficient to show that the
inference is invalid:
a b c d
T T T T
T T T F
F T T F
6. Sometimes there is no proposition which forces us into assigning truth values
to any of the proposition letters involved. In such cases we must try out different
assignments one at a time. Consider the following such inference:
a b d
b v d
d ⊃ a
b & a
In this inference we have a conclusion which is a conjunction together with
premises which are a disjunction and a conditional. We are not forced to assign T or F to
any letter. Instead, we must work through the various possible assignments. To make the
conclusion false, three assignments are possible: both "b" and "a" are false, or, "b" alone
is false, or, "a" alone is false. We can work through each of these. If none of them allow
for true premises and a false conclusion, the inference is valid.
a
F
T
F
b d
F
F
T
b v d
F
F
T
d ⊃ a
F
T
F
b
F
F
T
&
F
F
F
a
F
T
F
Making both false (on the first line), we see that "d" must be F if the second premise
is to be true, but when we do that, we see that the first premise is false. We get:
a
F
T
F
b d
F F
F
T
b v d
F F F
F
T
d ⊃ a
F T F
T
F
b
F
F
T
&
F
F
F
a
F
T
F
Note that the failure of this assignment does not show that the inference is valid.
Validity means that no assignment results in true premises and a false conclusion; all we
have shown so far is that we do not get true premises and a false conclusion when we
assign F to "a" and "b". Since this was only one of the assignments that would make the
conclusion false, we must try the others.
We know that in order to make the conclusion false, either "b" or "a" must be false,
so let us consider the second possibility, which assigns F to 'b' and T to "a":
a
F
T
F
b d
F F
F
T
b v d
F F F
F
T
d ⊃ a
F T F
T T
F
b
F
F
T
&
F
F
F
a
F
T
F
In order to make the first premise true, "d" must be T, and now we have found an
assignment which shows that the inference is invalid, one which makes the premises true
and the conclusion false:
a
F
T
F
b d
F F
F T
T
b v d
F F F
F T T
T
d ⊃ a
F T F
T T T
F
b
F
F
T
&
F
F
F
a
F
T
F
At this point we can stop, since any assignment which yields true premises and a
false conclusion shows that the inference is not valid. There might be additional
assignments which show that the inference is invalid, but one is sufficient to show that it
is not valid. (The third assignment also demonstrates invalidity, in fact.)
To repeat, the crucial point of this example is that the fact that the first assignment
we tried did not show the inference to be not valid was not sufficient to show that it was
valid. To show that an inference is valid we would need to try all of the possible
assignments that will either make the conclusion false or make one of the premises true.
Exercises
Part 1. For each inference, use a Targeted Truth Table to determine whether the inference
is valid or not valid.
Sample
1. e ⊃ f
2. ~(e & f)
--------3. f
e
f
F
F
F
e ⊃ f
F
F T F
~(e & f)
F
T F F F
f
F (step 1)
F (step 2)
Not valid, when e is F and f is F.
(1)
1. p ⊃ a
2. ~(a v s)
--------3. ~p
(2)
1. t ⊃ s
2. s ⊃ b
-----3. t ⊃ b
Part 2. For each inference, make a translation key, put the inference in standard form
using symbolic and use a Targeted Truth Table to determine whether the inference is valid
or invalid.
(3)
If we paint the plant with soapy water, the aphids will disappear. But it's not the case
that either the aphids will disappear, or the spider mites. So we will paint the plant
with soapy water.
(4) If Jack gets more training, he qualifies for the Special Ops unit. If Jack qualifies for
the Special Ops unit, he is shipping to Baghdad. So, if Jack is shipping to Baghdad,
he gets more training.
(5)
If Jack gets more training, he will qualify for the Special Ops unit. If Jack qualifies
for the Special Ops unit, he will immediately be shipped to Baghdad. So, either Jack
gets more training or he will not immediately be shipped to Baghdad.
Answers to Even Numbers
(2)
1. t ⊃ s
2. s ⊃ b
------3. t ⊃ b
t
T
T
s
b
F
F
t ⊃ s
T
T T T
s ⊃ b
F
T F F
t ⊃ b
T F F
T F F
Valid - It is impossible to assign values which make the inference invalid
(4)
If Jack gets more training, he qualifies for the Special Ops unit. If Jack qualifies for the
Special Ops unit, he is shipping to Baghdad. So, if Jack is shipping to Baghdad, he gets
more training.
1. t ⊃ q
2. q ⊃ b
-----3. b ⊃ t
t
F
F
q
T
b
T
T
t ⊃ q
F
F T T
q ⊃ b
T
T T T
b ⊃ t
T F F
T F F
Invalid - when "t" is F, "q" is T and "b" is T, the premises can be made true and the
conclusion false
7 The Truth Tree Method
1. Truth trees, like truth tables, will show that an inference is valid if it is valid, or
not valid if it isn't. They have the advantage of dealing more briefly than truth tables with
inferences involving a number of simple propositions. Their spatial/graphical nature also
makes them more intuitive for some people.
2. Truth trees are like targeted truth tables. We assume the premises are true and
the conclusion is false. If the inference is valid, we have just assumed a contradiction, and
this contradiction will show itself when we look at the simple propositions. On the other
hand, if the simple propositions can be consistent, the inference is invalid.
3. We begin the process by writing the premises and the negation of the conclusion
in a single vertical column. In the truth tree method we do not use "T" and "F" as
contradictory truth values; rather, we use the simple proposition letters and their
negations. Thus, to negate the conclusion is to assume that the conclusion itself is false.
This initial vertical column can be thought of as the trunk of the tree starting to
grow. The tree is growing downwards, by the way. It is upside-down. (Perhaps this
method should be called the "roots" method!)
For example, a simple Elim. inference would be set up like this:
a v b
~a
~b
Note that the conclusion "b" has been negated.
4. The next step is to "decompose" any complex propositions listed, except
negations of simple propositions. To decompose a proposition, we strike it out and extend
each branch of the tree downwards in accordance with the rules of decomposition. A full
list of the rules of decomposition will be given shortly.
In our Elim. example, neither the second nor the third line need to be decomposed,
since they are both negated simple propositions. Only the first line needs to be
decomposed. The rule for decomposing a disjunction is to "branch" into the left-hand
disjunct and the right-hand disjunct. (A complete set of rules is given below.) We strike
out the line decomposed.
a v b
~a
~b
a
b
5. After each decomposition, check for contradictions along each branch,
including the trunk. In our example, moving up from the bottom of the left-hand branch,
we see that it contains a contradiction between "a" on line 4 and "~a" on line 2, and the
right-hand branch contains a contradiction between "b" on line 4 and "~b" on line 3. We
place an "X" below closed branches.
a v b
~a
~b
a
b
X
X
6. Our Elim. example has closed all of its branches. It is therefore valid. If there
were any unclosed branches, we would continue the process of decomposition, one
proposition at a time, until all complex propositions have been decomposed. If all of the
propositions have been decomposed and there are still unclosed branches, the inference
is not valid.
7. Some decomposition rules are branching and some are non-branching. By
"branching" we mean that the proposition decomposes horizontally, indicating that there
are two ways in which the proposition could be true, while "non-branching" means that
the proposition decomposes vertically, indicating that the truth of the proposition
requires that both parts are true.
Conjunctions, for example, decompose vertically. Consider the inference from "a
& b" to "a". We would set this up as:
a & b
~a
Only the first line needs to be decomposed. Conjunctions decompose vertically,
extending the trunk of the tree. After decomposing line 1 and looking for contradictions,
our example looks as follows:
a & b
~a
a
b
X
There is only one "branch" (in fact, the tree is just a trunk) and it contains a
contradiction, between "a" on line 3 and "~a" on line 2. The inference is valid.
8. At the end of the process, either every branch will be closed, or not every branch
will be closed. If every branch is closed, we have shown that the assumption of invalidity
was contradicted, and so the inference is valid. Each branch that remains open, however,
illustrates a counter-example to the validity of the inference, which shows the inference
to be invalid.
Here is a demonstration of the invalidity of MC. Letting "S" and "T" be "a" and "b",
the propositions involved are "a ⊃ b", "b", "a", and so we write down:
a ⊃ b
b
~a
There is only one proposition requiring decomposition ("a ⊃ b"), which decomposes by
branching into "~a" and "b". We add these branches to the trunk on the next line below
and strike out the proposition being decomposed:
a ⊃ b
b
~a
~a
b
We check each branch for contradictions. There are none. It is possible for the premises
to be true and the conclusion false, when "~a" is true (or, "a" is false) and "b" is true. The
inference is declared not valid.
9. Here is the complete set of rules of decomposition. With "S" and "T" standing
for any proposition, whether simple or complex, the non-branching rules are as follows:
S & T
~(S v T)
S
T
~(S ⊃ T)
~S
~T
~~S
S
~T
S
and the branching rules are as follows:
~(S & T)
~S
~T
S v T
S
S ⊃ T
T
~S
T
Remember that as you decompose each proposition, you strike it out in your tree.
Some of these rules will be familiar from the truth tables above. "S & T"
decomposes into a non-branching "S" and "T" because "S & T" is true when both "S" and
"T" are true. (Compare with the truth table for "S & T", above.) "S v T" branches into "S"
and "T" because a disjunction is made true when either one of the disjuncts is true. (Again,
compare the truth table for "S v T" above.) "S ⊃ T" branches because a conditional is true
when either the antecedent is not true or the consequent is true. (Again, compare the truth
table for "S ⊃ T" above.) The rules for "~~S" can be understood by comparison with the
truth table in section 2 and the rule of double negation from the Method of Derivation
(see 6.1 of that chapter) and "~(S & T)" and "~(S v T)" can be understood by comparison
with De Morgan's rule (in 6.6.)
10. Decompose non-branching propositions before branching ones. This is
because, if a tree already has branches, any further decompositions must be applied to all
of the branches below it. We can prevent some unnecessary writing by extending the trunk
first, since the trunk is included in all branches. Consider the following inference, in
standard form:
1. a v ~b
2. ~a & c
-------3. ~b
We set up the truth tree as follows, writing "~~b" on the third line as the negation of the
conclusion.
a v ~b
~a & c
~~b
It's a good idea to decompose "~a & c" and "~~b" before the disjunction in the first line,
because they are non-branching while it is branching. After decomposing the second line,
we get:
a v ~b
~a & c
~~b
~a
c
We check for contradictions, but seeing none, proceed to decompose the third line
("~~b"). We cross out "~~b" on the third line and add "b" to the bottom of the tree, to get:
a v ~b
~a & c
~~b
~a
c
b
There are still no contradictions, so we proceed to decompose the first line:
a v ~b
~a & c
~~b
~a
c
b
a
~b
Again, we check for contradictions on each branch. The left-hand side contains a
contradiction between the "a" on the final line and the "~a" on the fourth line. So, we place
an "X" beneath the left-hand branch. The right-hand side contains a contradiction,
between the "~b" on the final line and the "b" on the sixth line. So, we place an "X" beneath
the right-hand branch.
a v ~b
~a & c
~~b
~a
c
b
a
~b
X
X
Since we have finished decomposing, and there are only these two branches, both of which
have been closed off, the inference is valid.
If we had decomposed line 1 first, the tree would have looked like this:
a v ~b
~a & c
~~b
a
~b
Further decompositions (of line 2 and line 3) would then have to be written under both
branches. After decomposing line 2:
a v ~b
~a & c
~~b
a
~a
c
X
~b
~a
c
And after decomposing line 3:
a v ~b
~a & c
~~b
a
~a
c
X
~b
~a
c
b
X
11. Any unclosed branch is sufficient to show that the inference is not valid. In the
following example, which is already in truth tree format, the first two lines are the
premises, while the third is the negation of the conclusion.
a ⊃ f
(a & c) v ~b
~~b
We begin by decomposing third line ("~~b") without branching, on line four. Both the
first and second lines will cause the tree to branch. Taking the first line first, we get:
a ⊃ f
(a & c) v ~b
~~b
b
~a
f
We look for contradictions along each branch, and find none. The second line is now
decomposed. The results of its decomposition must be placed at the bottom of each
branch below it. We look for contradictions and find that the second and fourth branches
show a contradiction, between "~b" and "b" on the fourth line.
a ⊃ f
(a & c) v ~b
~~b
b
~a
a & c
f
~b
X
a & c
~b
X
Finally, we decompose "a & c":
~a ⊃ f
(a & c) v ~b
~~b
b
~a
a & c
a
c
f
~b
X
a & c
a
c
~b
X
The first branch closes, since the "a" on the seventh line contradicts the "~a" of the fifth
line:
~a ⊃ f
(a & c) v ~b
~~b
b
~a
a & c
a
c
X
f
~b
X
a & c
a
c
~b
X
However, the third branch does not close, and so the inference is invalid, when "a" is
true", "b" is true, "c" is true, and "f" is true.
Exercises
Sample
Use the truth tree method to show that the following (an instance of CC) is valid:
1. a ⊃ b
2. ~b
-----3. ~a
a ⊃ b
~b
~~a
a
~a
X
b
X
(1) Use the truth tree method to show that the following (an instance of MC) is
invalid:
1. a ⊃ b
2. b
------3. a
(2) Use the truth tree method to show that the following (an instance of Simp.) is valid:
1. a & b
-----2. a
(3) Use the truth tree method to show that the following (an instance of Elim.) is valid:
1. a v (b ⊃ c)
2. ~a
-----------3. b ⊃ c
(4) Use the truth tree method to show that the following (an instance of DD) is valid:
1. a ⊃ b
2. c ⊃ d
3. ~b v ~d
---------4. ~a v ~c
(5) Use the truth tree method to determine whether the following inference is valid or
invalid.
1. (b v ~c) ⊃ d
2. ~c
-------------3. d
(6)
Consider the following inference:
If national elections deteriorate into television popularity contests, smooth-talking
morons will get elected. So clearly, smooth-talking morons won't get elected if the
elections don't deteriorate into television popularity contests.
(a) Make a translation key for it.
(b) Relative to the key, translate the inference into symbolic and put it in standard
form.
(c) Relative to what you have in standard form, make a truth tree for the inference.
(d) Relative to your truth tree, is the inference valid or invalid?
Answers to Even Numbers
(2) Use the truth tree method to show that the following (an instance of Simp.) is valid:
a & b
~a
a
b
X
(4) Use the truth tree method to show that the following (an instance of DD) is valid:
a ⊃
c ⊃
~b v
~(~a v
~~a
~~c
a
c
b
d
~d
~c)
~b
~a
X
(6)
~d
b
X
~a
X
b
~c
X
d
X
Consider the following inference:
If national elections deteriorate into television popularity contests, smooth-talking
morons will get elected. So clearly, smooth-talking morons won't get elected if the
elections don't deteriorate into television popularity contests.
(a) Make a translation key for it.
d = National elections deteriorate into television popularity contests.
e = Smooth-talking morons get elected.
(b) Relative to the key, translate the inference into symbolic and put it in standard form.
1. d ⊃ e
------
2. ~d ⊃ ~e
(c) Relative to what you have in standard form, make a truth tree for the inference.
d ⊃ e
~(~d ⊃ ~e)
~d
~~e
e
~d
e
(d) Relative to your truth tree, is the inference valid or invalid?
Not valid – neither branch closes off.
8 A Summary Of Truth Conditions — Truth Tables & Truth Trees
Truth Tables
~ S
S v T
S & T
S ⊃ T
F T
T F
T
T
F
F
T
T
F
F
T
T
F
F
T
T
T
F
T
F
T
F
T
F
F
F
T
F
T
F
T
F
T
T
T
F
T
F
Truth Trees
Non-branching:
S & T
S
T
~(S v T)
~S
~T
~(S ⊃ T)
S
~T
~~S
S
Branching:
~S
~(S & T)
~T
S v T
S
S ⊃ T
T
~S
T