PREDICATES AND QUANTIFIERS
Lecture # 4
1
TERMINOLOGY REVIEW

Proposition:
- A statement that is either true or false
- Must always be one or the other!

Example:
 “The sky is red”
 Not a proposition: x + 3 > 4

Boolean variable:
- A variable (usually p, q, r, etc.) that represents a
proposition.
2
Consider the following statements:
x > 3, x = y +3, x + y = z
We can make propositions out of such statements


A predicate is a property that is affirmed or denied
about the subject (in logic: we say “variables” or
“arguments”) of a statement.

“ x
subject
3
is greater than 3”
predicate
To write in a predicate logic:
“ x is greater then 3”

subject
predicate
We introduce a (functional) symbol for the predicate,
and put the subject as an argument (to the functional
symbol): P(x). The statement P(x) is also said to be
propositional function.
Examples:


•
•
•
•
4
Father(x): unary predicate
Brother(x, y): binary predicate
Sum(x, y, z): ternary predicate
P(x, y, z, …t): n-ary predicate
PROPOSITIONAL FUNCTIONS
Consider P(x) = x < 5
 P(x) has no truth values (x is not given a value)



P(1) is true
 The proposition 1 < 5 is true
P(10) is false
 The proposition 10 < 5 is false
Thus, P(x) will create a proposition when given a value.

5
EXAMPLE
Let P(x) denote the statement “x > 3”. What are truth
values of P(4) and P(2)?
SOLUTION:
P(x) = x > 3
P(4) = 4 > 3
P(4) = True


P(x) = x > 3
P(2) = 2 > 3
P(2) = False
6
EXAMPLES
Let P(x) = “x is a multiple of 5”


For what values of x is P(x) true?
Let P(x) = x+1 > x


7
For what values of x is P(x) true?
PROPOSITIONAL FUNCTIONS
INVOLVING MULTIPLE VARIABLES
Functions with multiple variables:


P(x,y) = x + y == 0


P(x,y,z) = x + y == z

9
P(1,2) is false, P(1,-1) is true
P(3,4,5) is false, P(1,2,3) is true
DEFINITION

A predicate is a sentence that contains a finite number of
variables and becomes a statement when specific values
are substituted for the variables.

The domain of a predicate variable is the set of all values
that may be substituted in place of the variable.
10
UNIVERSE OF DISCOURSE

Consider the statement ‘x > 3’, does it make sense to
assign to x the value ‘blue’?

Intuitively, the universe of discourse is the set of all
things we wish to talk about; that is the set of all objects
that we can sensibly assign to a variable in a propositional
function. Referred to as Domain.

What would be the universe of discourse for the
propositional function below be:
EnrolledCSE235(x)=‘x is enrolled in CSE235’
11
QUANTIFIERS

When the variables in a propositional function are
assigned values, the resulting statement becomes a
proposition with a certain truth values.

However, there is another important way, called
quantification, to create a proposition from a
propositional function.

Quantification expresses the extent to which a
predicate is true over a range of elements.
12

In English, the words all, some, many, none and few are
used in quantification.

We will focus on two types of quantification here
 Universal Quantification 
 Existential Quantification 
13

The statement ‘x > 3’ is not a proposition

It becomes a proposition



When we assign values to the argument: ‘4>3’ is false, ‘2<3’ is true, or
When we quantify the statement
Two quantifiers

Universal quantifier 
\for all
the proposition is true for all possible values in the universe of discourse

Existential quantifier 
\exists
the proposition is true for some value(s) in the universe of discourse
14
UNIVERSAL QUANTIFIER

Definition: The universal quantification of a predicate P(x) is
the proposition ‘P(x) is true for all values of x in the universe of
discourse.’

Notation:
We use the notation:  x P(x), which is read ‘for all x’.

If the universe of discourse is finite, say {n1,n2,…,nk}, then the
universal quantifier is simply the conjunction of the
propositions over all the elements
 x P(x)  P(n1)  P(n2)  …  P(nk)
15
UNIVERSAL QUANTIFIER

Represented by an upside-down A: 



It means “for all”
Let P(x) = x+1 > x
We can state the following:



16
x P(x)
English translation: “for all values of x, P(x) is true”
English translation: “for all values of x, x+1>x is true”
EXAMPLE

Let the universe be the real numbers.


Then, x P(x) is true
Let P(x) = x/2 < x


Not true for the negative numbers!
Thus, x P(x) is false

When the domain is all the real numbers

In order to prove that a universal quantification is true, it
must be shown for ALL cases.

In order to prove that a universal quantification is false, it
must be shown to be false for only ONE case
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LOOP

Think of  as a for loop:

x P(x), where 1 ≤ x ≤ 10

… can be translated as …
for ( x = 1; x <= 10; x++ )
is P(x) true?

If P(x) is true for all parts of the for loop, then x P(x)

18
Consequently, if P(x) is false for any one value of the for loop, then
x P(x) is false
REMEMBER

The truth value of  x P(x) depends on the
domain/universe of discourse!

A statement  x P(x) is false. If and only If P(x) is not
always true when x is in the domain.

One way to show that P(x) is not always true when x is
in the domain is to find a counterexample.

A single counter example is all we need to establish
that  x P(x) is false.
19
EXAMPLE

Let Q(x) be the statement “x < 2”. What is the truth
value of the quantification  x Q(x), where the domain
consists of all real numbers?

Solution:
Q(x) is not true for every real number x
Counter Example:
Let x = 3
Q(3) is false
Thus,
 x Q(x)
is false.
20
EXAMPLE

What is the truth value of  x P(x), where P(x) is the statement
“x2 < 10” and the domain consists of the positive integers not
exceeding 4?

Solution:
The statement  x P(x) is the same as the conjunction
P(1)  P(2)  P(3)  P(4)
P(1) is true
P(2) is true
P(3) is true
P(4) is false
P(4), which is the statement “42 < 10” is false,
Thus,
 x P(x) is false.
21
EXAMPLE

a) What is the truth value of  x (x2 ≥ x), if the domain
consists of all real numbers?
b) What is the truth value of  x (x2 ≥ x), if the domain
consists of all integers?
a) Solution:
The universal quantification (x2 ≥ x) is not true for
every real number x
Counter Example:
Let x = ½
(1/2)2 ≥ (1/2) is false
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b) Solution:

The universal quantification (x2 ≥ x) is true for
all integers.

The universal quantification (x2 ≥ x) is true for
all 0 < x < 1.
23
EXISTENTIAL QUANTIFICATION

Definition: The existential quantification of a predicate P(x) is
the proposition ‘There exists a value x in the universe of
discourse such that P(x) is true.’

Notation:
We use the notation:  x P(x), which is read ‘there exists x’.

If the universe of discourse is finite, say {n1,n2,…,nk}, then the
existential quantifier is simply the disjunction of the
propositions over all the elements
 x P(x)  P(n1)  P(n2)  …  P(nk)
24

Represented by an backwards E: 



It means “there exists”
Let P(x) = x+1 > x
We can state the following:



25
x P(x)
English translation: “there exists (a value of) x such that P(x) is
true”
English translation: “for at least one value of x, x+1>x is true”
EXAMPLE

Let P(x) = x+1 > x
 There is a numerical value for which x+1>x
 In fact, it’s true for all of the values of x!
 Thus, x P(x) is true

In order to show an existential quantification is true, you
only have to find ONE value

In order to show an existential quantification is false, you
have to show it’s false for ALL values
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EXAMPLE

Let P(x) be the statement “x > 3.” What is the truth value
of the quantification  x P(x), where the domain consists of
all real numbers?

Solution:
P(x) is true for some values in domain
Counter Example:
Let x = 4
P(4) is true
Thus,
 x P(x)
is true.
27
EXAMPLE

Let Q(x) be the statement “x = x + 1.” What is the truth
value of the quantification  x Q(x), where the domain
consists of all real numbers?

Solution:
Q(x) is false for every real number x.
Counter Example:
Let x = 6
Q(6) = 6 = 6 + 1 is false
Thus,
 x Q(x)
is false.
28
EXAMPLE

What is the truth value of  x P(x), where P(x) is the statement
“x2 > 10” and the universe of discourse consists of the positive
integers not exceeding 4?

Solution:
The statement  x P(x) is the same as the disjunction
P(1)  P(2)  P(3)  P(4)
P(1) is false
P(2) is false
P(3) is false
P(4) is true
P(4), which is the statement “42 < 10” is true
Thus,
 x P(x) is true
29
NOTE

Recall that P(x) is a propositional function


Recall that a proposition is a statement that is either true
or false


Let P(x) be “x == 0”
P(x) is not a proposition
There are two ways to make a propositional function into
a proposition:

Supply it with a value


For example, P(5) is false, P(0) is true
Provide a quantification

For example, x P(x) is false and x P(x) is true

30
Let the universe of discourse be the real numbers
QUANTIFIERS: Truth values

In general, when are quantified statements true or false?
Statement True when…
x P(x)
x P(x)
False when...
There is an x for which
P(x) is true for every x
P(x) is false
There is an x for which
P(x) is false for every x
P(x) is true
x P(x) is true when domain is empty.
x P(x) is false when domain is empty.
UNIQUENESS QUANTIFIER

The Uniqueness quantifier, denoted by ! or 1.

Notation:

! x P(x) [or 1 x P(x)] states “there exist a unique x such that
P(x) is true”

Other phrases for uniqueness quantification include “there is
exactly one” and “there is one and only one.”
32
EXAMPLE

! x (x - 1 = 0), where the domain is the set of real

Solution:
(x - 1 = 0) is true for a unique real number.
Counter Example:
Let x = 1
(x - 1 = 0) is true
Note:
Generally, uniqueness quantifier is avoided.
It is best to stick with existential and universal quantifier.
numbers.
33
QUANTIFIERS WITH RESTRICTED
DOMAINS

EXAMPLE:
What do statements  x < 0 (x2 > 0),
 y ≠ 0 (y3 ≠ 0)
SOLUTION:
The statement  x < 0 (x2 > 0) states that for every real
number x with x < 0, x2 > 0.
That is, “the square of negative real number is positive.”
The statement is the same as:
 x (x < 0 → x2 > 0).
34

The statement  y ≠ 0 (y3 ≠ 0) states for every real
number y with y ≠ 0, we have (y3 ≠ 0).
That is, “the cube of every nonzero real number is
nonzero.”
This statement is equivalent to:
y(y ≠ 0 → y3 ≠ 0)
35
NOTE:

The restriction of a universal quantification is the same as the
universal quantification of a conditional statement.
Example:
 x < 0 (x2 > 0) is another way of expressing
x (x < 0 → x2 > 0)
The restriction of a existential quantification is the same as the
existential quantification of a conjunctive statement.
Example:
False
 z > 0 (z2 = 2) is another
way of expressing
 z (z > 0 ^ z2 = 2)

36
PRECEDENCE OF QUANTIFICATION

The quantifiers  and  have higher precedence than all
logical operators from propositional calculus.

Example:
 x P(x)  Q(x) is the disjunction of  x P(x) and Q(x).


In other words,
It means
( x P(x))  Q(x) rather than  x (P(x)  Q(x)).
37
BINDING VARIABLES

When a quantifier is used on the variable x, we say that
the occurrence of the variable is bound.

An occurrence of a variable that is not bound by a
quantifier is said to be free.

The part of a logical expression to which a quantifier is
applied is called the scope of this quantifier.
38
EXAMPLE

In the statement  x (x + y =1),

The variable x is bound to by the existential
quantification  x.

The variable y is free because it is not bound by a quantifier.
 x (x + y =1), x is bound, but y is free.
39
EXAMPLE

In the statement x (P(x)  Q(x))  x R(x), all variables
are bound.

The scope of the first quantifier, x, is the expression
(P(x)  Q(x))
The scope of the second quantifier, x is the expression
R(x)
That is, the existential quantifier binds the variable x in
P(x)  Q(x)
Universal quantifier binds the variable x in R(x).



40
EXAMPLES

(x P(x))  Q(x)


(x P(x))  (x Q(x))


Both x values are bound; thus it is a proposition
(x (P(x)  Q(x)))  (y R(y))


The x in Q(x) is not bound; thus not a proposition
All variables are bound; thus it is a proposition
(x P(x)  Q(y)  (y R(y))

41
The y in Q(y) is not bound; this not a proposition
BINDING VARIABLES

Let P(x, y) be x > y

Consider: x P(x, y)



This is not a proposition!
What is y?
 If it’s 5, then x P(x, y) is false
 If it’s x-1, then x P(x, y) is true
Note that y is not “bound” by a quantifier
42
LOGICAL EQUIVALENCES INVOLVING
QUANTIFIERS

Two statements S and T involving predicates and
quantifiers are logically equivalent if and only if they have
the same truth value regardless of the interpretation,
i.e. regardless of




The meaning that is attributed to each propositional function.
The domain of discourse.
Notation:
We use the notation S  T to indicate that two
statements S and T involving predicates and quantifiers
are logically equivalent.
43
EXAMPLE

Is x (P(x)  Q(x)) logically equivalent to x P(x)  x Q(x)?
where the same domain is used throughout.

Solution:
Use two steps


If x (P(x)  Q(x)) is true, then x P(x)  x Q(x) is true

Proof: Suppose x (P(x)  Q(x)) is true.
Then if a is in the domain, P(a)  Q(a) is true,
So P(a) is true and Q(a) is true.

So, if a is in the domain P(a) is true, which is the same as x P(x) is true;
and similarly, we get that x P(x) is true
44
TRANSLATING FROM ENGLISH INTO
LOGICAL EXPRESSIONS



Example: Express the statement using predicates and
quantifiers.
“Every student in this class has studied calculus”
Solution:
Step 1: First we rephrase the statement so that we can
clearly identify the appropriate quantifiers to use.
“For every student in this class, that student has studied
calculus.”
Step 2: Introduce a variable x
“For every student x in this class, x has studied calculus”
45

Step 3: Introduce C(x)
C(x) = x has studied calculus.
Domain = Students in the class.
Step 4:
x C(x)
46
EXAMPLE

Express the statement using predicates and quantifiers.
“For every person x, If person x is a student in this class then x
has studied calculus.”




Solution:
Step 1:
Step 2:
Step 3: Introduce S(x) and C(x)
S(x) = Person x is in the class.
C(x) = x has studied Calculus.
Domain = Students in the class.
Step 4:
x (S(x) → C(x))
47
EXAMPLE

Express the statement using predicates and quantifiers.
“Some student in this class has visited Mexico”
Solution:
 Step 1: First we rephrase the statement so that we can
clearly identify the appropriate quantifiers to use.
“There is a student in this class with the property that the
student has visited Mexico”

Step 2: Introduce a variable x
“There is a student x in this class having the property that
x has visited Mexico”
48

Step 3: Introduce M(x)
M(x) = x has visited Mexico.
Domain = Students in the class.
Step 4:
 x M(x)
49
EXAMPLE

Express the statement using predicates and quantifiers.
“There is a person x having the properties that x is a student
in this class and x has visited Mexico.”




Solution:
Step 1:
Step 2:
Step 3: Introduce S(x) and M(x)
S(x) = x is a student in this class.
M(x) = x has visited Mexico.
Domain = All person.
Step 4:
 x (S(x)  M(x))
50
EXAMPLE

Express the statement using predicates and quantifiers.
“For every x in this class, x has the property that x has visited
Mexico or x has visited Canada.”




Solution:
Step 1:
Step 2:
Step 3: Introduce C(x) and M(x)
C(x) = x has visited Canada.
M(x) = x has visited Mexico.
Domain = All people.
Step 4:
 x (C(x)  M(x))
51
EXAMPLE

Express the statement using predicates and quantifiers.
“For every person x, if x is a student in this class, then x has visited
Mexico or x has visited Canada.”




Solution:
Step 1:
Step 2:
Step 3: Introduce S(x) and M(x)
S(x) = x is a student in this class.
C(x) = x has visited Canada.
M(x) = x has visited Mexico.
Domain = student in the class.
Step 4:
 x (S(x) → (C(x)  M(x))).
52
EXAMPLES FROM LEWIS CARROL

Consider these statements. The first two are called
premises and the third is called the conclusion. The entire
set is called an argument.
“All lions are fierce.”
“Some lions do not drink coffee.”
Therefore,
“Some fierce creatures do not drink coffee.”
53
SOLUTION

Let
P(x) = “x is a lion.”
Q(x) = “x is fierce.”
R(x) = “x drinks coffee.”
Domain = Assume all creatures.


We can express these statements as:
“All lions are fierce.”
 x (P(x) → Q(x)).
54

“Some lions do not drink coffee.”
x (P(x)  ¬R(x)).

“Some fierce creatures do not drink coffee.”
x (Q(x)  ¬R(x)).
55
MORE EXAMPLES

Translate the statements:





“All hummingbirds are richly colored”
“No large birds live on honey”
“Birds that do not live on honey are dull in color”
“Hummingbirds are small”
Solution:
Assign propositional functions
Let
P(x) = “x is a hummingbird”
Q(x) = “x is large”
R(x) = “x lives on honey”
S(x) = “x is richly colored”
Let our universe of discourse be all birds
56


Translate the statements:
Assume that “small” is the same as “not large” and that
“dull in color” is the same as “not richly colored”

“All hummingbirds are richly colored”
 x (P(x)→S(x))

“No large birds live on honey”
 ¬x (Q(x)  R(x))
 Alternatively: x (¬Q(x)  ¬R(x))

“Birds that do not live on honey are dull in color”
 x (¬R(x) → ¬S(x))

“Hummingbirds are small”
 x (P(x) → ¬Q(x))
57
NEGATING QUANTIFIED
EXPRESSIONS

Consider the statement:
“Every student in your class has taken a course in
calculus.”
P(x) = x has taken a course in calculus.
Domain = Students in your class.
x P(x)
58

The negation of this statement is
“It is not the case that every student in your class has
taken a course in calculus.”

This is equivalent to
“There is a student in your class who has not taken a
course in calculus.”
x ¬P(x)
59

This example illustrate the following logical equivalence:
¬x P(x)  x ¬P(x)
60
NOTE

First note that ¬x P(x) is true if and only if x P(x) is
false.

Note that x P(x) is false if and only if there is an
element x in domain for which P(x) is false.

Finally, note that there is an element x in the domain for
which ¬ P(x) is true if and only if x ¬P(x) is true.

Putting these steps together, we can conclude that:
¬x P(x) is true if and only if x ¬P(x) is true.
61

It follows that ¬x P(x) and x ¬P(x) are logically
equivalent.
62

Another example illustrate that following logical
equivalence:
¬x Q(x)  x ¬Q(x)
63
EXAMPLE

What are the negations of the statements:
“There is an honest politician”

Solution:
H(x) = x is an honest politician.
Domain = all politician
x H(x)
The negation of this statement is
¬(x H(x)) = x ¬H(x)
“Every politician is dishonest.”
64
EXAMPLE

What are the negations of the statements:
“All Americans eat cheeseburgers”

Solution:
C(x) = x eats cheeseburgers.
Domain = all Americans
x C(x)
The negation of this statement is
¬(x C(x)) = x ¬C(x)
“Some Americans does not eat cheeseburgers.”
65
EXAMPLE

What are negations of the statement:
x (x2 > x)

Solution:
The negation of x (x2 > x) is the statement ¬x (x2 > x)
¬x (x2 > x) = x ¬(x2 > x)
This can be re-written as:
x (x2 ≤ x)
66
EXAMPLE

What are negations of the statement:
x (x2 = 2)

Solution:
The negation of x (x2 = 2) is the statement ¬x (x2 = 2)
¬x (x2 = 2) = x ¬(x2 = 2)
This can be re-written as:
x (x2 ≠ 2)
67
EXMPLE

Show that ¬x (P(x)→Q(x)) and x (P(x)  ¬Q(x)) are
logically equivalent.

Solution:
 ¬x (P(x)→Q(x))
 x ¬(P(x)→Q(x))
Given
De Morgan’s law for universal
quantifiers
 x ¬(¬P(x)  Q(x)) As we know that p → q = ¬p  q
 x (¬(¬P(x))  ¬Q(x)) De-Morgan’s law
 x (P(x)  ¬Q(x))
Double Negation law
 R.H.S
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EXMPLE

Express the negation of xyP(x,y)  xyQ(x,y).

Solution:
 ¬(xyP(x,y)  xyQ(x,y))
 ¬(xyP(x,y))  ¬(xyQ(x,y))
 x¬(yP(x,y))  x¬(yQ(x,y) )
 xy ¬P(x,y)  xy ¬Q(x,y)
69
Given
De Morgan’s law
De Morgan’s law
De-Morgan’s law
RULES OF INFERENCE FOR
QUANTIFIED STATEMENTS
70
INTRODUCTION

We have discussed rules of inference for propositions.
We will now describe some important rules of inference
for statements involving quantifiers, which are given
below:




71
Universal Instantiation
Universal Generalization
Existential Instantiation
Existential Generalization
UNIVERSAL INSTANTIATION

Assume that we know that x P(x) is true:

Then we can conclude that P(c) is true




Here c stands for some specific constant from the domain.
This is called “universal instantiation”
Given the premise x P(x).
Example:
We have a statement “All woman are wise.”
We can conclude that “Sophia is wise”
Where Sophia is member of the domain.
72
UNIVERSAL GENERALIZATION

Assume that we know that P(c) is true for any value of c


73
Then we can conclude that x P(x) is true.
This is called “universal generalization”
EXISTENTIAL INSTANTIATION

Assume that we know that x P(x) is true



74
Then we can conclude that P(c) is true for some value of c
 Here c stands for some specific constant from the
domain.
This is called “existential instantiation”
Given the premise x P(x).
EXISTENTIAL GENERALIZATION

Assume we know that P(c) is true for some value of c


75
Then we can conclude that x P(x) is true
This is called “existential generalization”
76
EXAMPLE

Show that the premises:
“Everyone in this discrete structures class has taken a
course in computer science” and
“John is a student in this class” imply the
conclusion “John has taken a course in computer
science.”

Solution:
D(x) = x is in discrete mathematics class.
C(x) = x has taken a course in computer science.
77

Then the premises are:
x (D(x) → C(x))
D(John)
C(John)

The following steps can be used
to establish conclusion:
1. x (D(x) → C(x))
2. D(John) → C(John)
3. D(John)
4.
C(John)

Premise
Universal
Instantiation
Premise
Modus Ponens
Rule of inference
pq
p
q
q
pq
 p
pq
qr
Tautology
Name
[ p  ( p  q)]  q
Modus ponens
[q  ( p  q)]  p
Modus tollen
[( p  q)  (q  r )]  ( p  r ) Hypothetic al syllogism
pr
pq
p
q
p
pq
pq
p
p
q
pq
pq
p  r
q  r
(( p  q)  p)  q
Disjunctiv e syllogism
p  ( p  q)
Addition
( p  q)  p
Simplifica tion
(( p)  (q))  ( p  q)
Conjunctio n
[( p  q)  (p  r )]  ( p  r )
Resolution
EXAMPLE

Show that the premises:
“A student in this class has not read the book.” and
“Everyone in this class passed the first exam” imply the
conclusion “Someone who passed the first exam has not
read the book.”

Solution:
C(x) = x is in this class.
B(x) = x has read the book.
P(x) = x passed the first exam.
79
Then the premises are:
 x (C(x)  ¬B(x))
x (C(x) → P(x))
 x (P(x)  ¬B(x))
 The following steps can be used
to establish conclusion:
Rule of inference
1.  x (C(x)  ¬B(x))
Premise
pq
p
2. C(a)  ¬B(a)
Existential Instantiation
q
q
3. C(a)
Simplification
pq
 p
4. x (C(x) → P(x))
Premise
pq
qr
5. C(a) → P(a)
Universal Instantiation
pr
pq
6. P(a)
Modus Ponens
p
7. ¬B(a)  C(a)
Commutative on (2)
q
p
8. ¬B(a)
Simplification
pq
pq
9. P(a)  ¬B(a)
Conjunction on (6) & (8)  p
p
10.  x (P(x)  ¬B(x))
Existential Generalization q


pq
pq
p  r
q  r
Tautology
Name
[ p  ( p  q )]  q
Modus ponens
[q  ( p  q )]  p
Modus tollen
[( p  q )  (q  r )]  ( p  r ) Hypothetic al syllogism
(( p  q )  p )  q
Disjunctiv e syllogism
p  ( p  q)
Addition
( p  q)  p
Simplifica tion
(( p )  (q ))  ( p  q )
Conjunctio n
[( p  q )  (p  r )]  ( p  r )
Resolution
COMBINING RULES OF INFERENCE FOR
PROPOSITIONS & QUANTIFIED STATEMENTS

A rule of inference for quantified statements, and modus
ponens, a rule of inference for propositional logic.

We will often need to use this combination of rules of
inference. Because universal instantiation and modus
ponens are used so often together.

This combination of rules is sometimes called universal
modus ponens.
81
UNIVERSAL MODUS PONENS

This rule tells us that if x (P(x) → Q(x)) is true, and if
P(a) is true for a particular element a in the domain of
the universal quantifier, then Q(a) must also be true.

To see this, note that by universal instantiation
P(a) → Q(a) is true.
x (P(x) → Q(x))
P(a), where a is a particular element in the domain.
Q(a)
82
NOTE

Universal modus ponens is commonly used in
mathematical arguments.
83
UNIVERSAL MODUS TOLLENS

Universal modus tollens combines universal instantiation
and modus tollens and can be expressed in the following
way:
x (P(x) → Q(x))
¬Q(a), where a is a particular element in the domain.
 ¬P(a)
84
EXERCISE

Use rules of inference to show that if
x (P(x) → (Q(x)  S(x))) and
x (P(x)  R(x))) are true
then x (R(x)  S(x)) is true

Solution:
85
The following steps can be used to
establish conclusion:
01. x (P(x)  R(x))) Premise
02. P(a)  R(a)
Universal Instantiation
03. P(a)
Simplification
04. x (P(x) → (Q(x)  S(x))) Premise
05. Q(a)  S(a)
Universal modus ponens
06. S(a)  Q(a)
Commutative
07. S(a)
Simplification

Rule of inference
pq
08. R(a)  P(a)
09. R(a)
10. R(a)  S(a)
11. x (R(x)  S(x))
Commutative on (2)
Simplification
Conjunction (9) & (7)
Universal generalization
p
q
q
pq
 p
pq
qr
Tautology
Name
[ p  ( p  q )]  q
Modus ponens
[q  ( p  q )]  p
Modus tollen
[( p  q )  (q  r )]  ( p  r ) Hypothetic al syllogism
pr
pq
p
q
p
pq
pq
p
p
q
pq
pq
p  r
q  r
(( p  q )  p )  q
Disjunctiv e syllogism
p  ( p  q)
Addition
( p  q)  p
Simplifica tion
(( p )  (q ))  ( p  q )
Conjunctio n
[( p  q )  (p  r )]  ( p  r )
Resolution