Discrete Mathematics: Lecture 2 - Predicate Logic

Math211
Discrete Mathematics
Predicate Logic
SAHAR SELIM

Agenda
 1.4 Predicates and Quantifiers
 1.5 Nested Quantifiers
Sahar Selim MATH211 Lecture 3 | Predicate Logic
2

1.4 Predicates and Quantifiers
3

Predicate Logic
 A predicate is a proposition that is a function of one or
more variables in a given domain.
 E.g. P(x): x is an even number.
So P(1) is false, P(2) is true,….
 Examples of predicates:
 Domain ASCII characters - IsAlpha(x): TRUE iff x is an alphabetical
character.
 Domain floating point numbers - IsInt(x): TRUE iff x is an integer.
 Domain integers: Prime(x) - TRUE if x is prime, FALSE otherwise.
4

Subjects and Predicates
 In the sentence “The dog is sleeping”:
 The phrase “the dog” denotes the subject - the object or entity that the
sentence is about.
 The phrase “is sleeping” denotes the predicate- a property that is true of
the subject.
 In predicate logic, a predicate is modeled as a function P( )
from objects to propositions.
 P(x) = “x is sleeping” (where x is any object).
5

More About Predicates
 Convention:
 Lowercase variables x, y, z... denote objects/entities.
 Uppercase variables P, Q, R… denote propositional functions (predicates).
 Keep in mind that the result of applying a predicate P to an object x is
the proposition P(x).
 But the predicate P itself (e.g. P=“is sleeping”) is not a proposition (not a
complete sentence).
 E.g. if P(x) = “x is a prime number”,
P(3) is the proposition “3 is a prime number.”
6

Example
Let Q(x, y) denote the statement “x = y + 3”
What are the truth values of the propositions
Q(1, 2) and Q(3, 0)?
 When the variables in a propositional function are
assigned values, the resulting statement becomes a
proposition with a certain truth value.
7

Propositional Functions
 Predicate logic generalizes the grammatical notion of
a predicate to also include propositional functions of
any number of arguments, each of which may take
any grammatical role that a noun can take.
 E.g. let P(x,y,z) = “x gave y the grade z”, then
If x=“Mike”, y=“Mary”, z=“A”,
then P(x,y,z) = “Mike gave Mary the grade A.”
8

Quantifiers
9

Quantifiers
 In English, the words all, some, many, none, and few are
used in quantifications.
 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.
 It describes the values of a variable that make the predicate true.
E.g. ∃x P(x)
 Domain or universe: range of values of a variable
10

Universe of Discourse (U.D.s)
 The power of distinguishing objects from predicates is that it lets
you state things about many objects at once.
 E.g., let P(x)=“x+1>x”. We can then say,
“For any number x, P(x) is true” instead of
(0+1>0) ∧ (1+1>1) ∧ (2+1>2) ∧ ...
 The collection of values that a variable x can take is called x’s
Universe of Discourse / Domain of Discourse.
 Quantifiers provide a notation that allows us to quantify (count)
how many objects in the univ. of disc. satisfy a given predicate.
11

Two Popular Quantifiers
1. Universal: ∀x P(x) – “P(x) for all x in the domain”
2. Existential: ∃x P(x) – “P(x) for some x in the domain” or “there exists
x (that is, 1 or more) such that P(x) is TRUE”.
 Either is meaningless if the domain is not known/specified.
 Examples (domain real numbers)
 ∀ x (x2 >= 0)
 ∃ x (x >1)
 (∀x>1) (x2 > x) – quantifier with restricted domain
12

The Universal Quantifier ∀
 Example:
Let the u.d. of x be parking spaces at Cairo University.
Let P(x) be the predicate “x is full.”
Then the universal quantification of P(x), ∀x P(x), is the proposition:
 “All parking spaces at CU are full.”
 i.e., “Every parking space at CU is full.”
 i.e., “For each parking space at CU, that space is full.”
13

The Existential Quantifier ∃
 Example:
Let the u.d. of x be parking spaces at CU.
Let P(x) be the predicate “x is full.”
Then the existential quantification of P(x), ∃x P(x), is the proposition:
 “Some parking space at CU is full.”
 “There is a parking space at CU that is full.”
 “At least one parking space at CU is full.”
14

Example
 P(x) denote a predicate in the domain U = {1, 2, 3}
∀x P(x) ⇔ P(1) ∧ P(2) ∧ P(3)
∃x P(x) ⇔ P(1) ˅ P(2) ˅ P(3)
15

Truth of Quantifiers
16

Using Quantifiers
Domain integers:
 Using implications:
The cube of all negative integers is negative.
∀x (x < 0) →(x3 < 0)
 Expressing sums:
n
∀n (Σ i = n(n+1)/2)
i=1
17

Precedence of Quantifiers and
Logical Operators
Operator Precedence
∀ , ∃ 1
¬ 2
∧ 3
∨ 4
→ 5
↔ 6
18

Precedence of Quantifiers
 ∀ ∃ have higher precedence than operators from Propositional
Logic; so
 ∀x P(x) ∨ Q(x) is not logically equivalent to ∀x (P(x) ∨ Q(x))
 ∃x (P(x) ∧ Q(x)) ∨ ∀x R(x)
Say P(x): x is odd, Q(x): x is divisible by 3, R(x): (x=0) ∨(2x >x)
 Logical Equivalence: P ≡ Q iff they have same truth value no matter
which domain is used and no matter which predicates are assigned
to predicate variables.
19

Free and Bound Variables
 An expression like P(x) is said to have a free variable x (meaning,
x is undefined).
 A quantifier (either ∀ or ∃) operates on an expression having one
or more free variables, and binds one or more of those variables,
to produce an expression having one or more bound variables.
20

Remember
A predicate is not a proposition until all variables
have been bound either by quantification or
assignment of a value!
21

Binding Variables
 The occurrence of a variable x is bound iff a value is
assigned to x or when a quantifier is used on x.
Otherwise, it is called free.
22
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Example of Binding
 P(x,y) has 2 free variables, x and y.
 ∀x P(x,y) has 1 free variable, and one bound variable.
[Which is which?]
 “P(x), where x=3” is another way to bind x.
 An expression with zero free variables is an actual proposition.
 An expression with one or more free variables is still only a
predicate: ∀x P(x,y)
y x
23

Examples
 Let U = Z, the integers = {. . . -2, -1, 0 , 1, 2, 3, . . .}
 P(x): x > 0 is the predicate. It has no truth value until the variable x is bound.
 Examples of propositions where x is assigned a value:
 P(-3) is false,
 P(0) is false,
 P(3) is true.
 The collection of integers for which P(x) is true are the positive
integers.
24

Examples
 P(y) ˅ ¬P(0) is not a proposition.
 The variable y has not been bound.
 However, P(3) ˅ ¬P(0) is a proposition which is true.
 Let R be the three-variable predicate R(x, y z):
x + y = z
 Find the truth value of
R(2, -1, 5), R(3, 4, 7), R(x, 3, z)
25

Negation of Quantifiers
Negation of “All Egyptians love falafel”
“Not All Egyptians love falafel” ≡ “There exists some Egyptians
that do not love falafel”
¬ ∀x P(x) ≡ ∃x ¬P(x)
¬ ∃x P(x) ≡ ∀x ¬P(x)
 Careful: The negation of “Every Canadian loves Hockey” is NOT
“No Canadian loves Hockey”!
26

Negation of Quantifiers
 Distributing a negation operator across a quantifier
changes a universal to an existential and vice versa.
¬(∀x P(x)) ⇔ ¬(P(x1) ∧ P(x2) ∧ … ∧ P(xn))
⇔ ¬P(x1) ˅ ¬P(x2) ˅ … ˅ ¬P(xn)
⇔ ∃x ¬P(x)
27

De Morgan’s Laws for Quantifiers
28

Example
 What are the negations of the statements “There is an honest
politician” and “All Americans eat cheeseburgers”?
Solution:
 H(x) denote “x is honest.”
 “There is an honest politician” is represented by ∃xH(x), where
the domain consists of all politicians.
 Negation of this statement is ￢∃xH(x) ≡ ∀x￢ H(x).
29

Example
 Let C(x) denote “x eats cheeseburgers.”
 Then “All Americans eat cheeseburgers” is represented by
∀xC(x), where the domain consists of all Americans.
 The negation of this statement is ￢∀xC(x), which is equivalent to
∃x￢C(x).
 This negation can be expressed in several different ways,
including “Some American does not eat cheeseburgers” and
“There is an American who does not eat cheeseburgers.”
30

1.5 Nested Quantifiers
33

Nested Quantifiers
 Use distinct variables
 ∀x (∃y(P(x,y) → ∀x Q(y, x)))
 Variable name doesn’t matter
 ∀x ∃y P(x, y) ≡ ∀a ∃b P(a, b)
 Positions of quantifiers can change (but order is
important)
 ∀x (Q(x) ∧ ∃y P(x, y)) ≡ ∀x ∃y (Q(x) ∧ P(x, y))
34

Nested Quantifiers
Example: Let the u.d. of x & y be people.
Let L(x,y)=“x likes y” (a predicate with 2 free variables)
Then ∃y L(x,y) = “There is someone whom x likes.” (A predicate with
1 free variable, x)
Then ∀x (∃y L(x,y)) = “Everyone has someone whom they like.”
(A __________ with ___ free variables.)
35

∀x ∃y (x + y = 0) is true over the integers
 Assume an arbitrary integer x.
 To show that there exists a y that satisfies the requirement of the
predicate, choose y = -x. Clearly y is an integer, and thus is in the
domain.
 So x + y = x + (-x) = x – x = 0.
 Since we assumed nothing about x (other than it is an integer), the
argument holds for any integer x.
 Therefore, the predicate is TRUE.
Nested Quantifiers
36

 Caution: In general, order matters! Consider the following
propositions over the integer domain:
∀x ∃y (x < y) and ∃y ∀x (x < y)
 ∀x ∃y (x < y) : “there is no maximum integer”
 ∃y ∀x (x < y) : “there is a maximum integer”
 Not the same meaning at all!!!
Nested Quantifiers
37

Quantification as loop
 For every x, for every y: ∀x ∀y P(x, y)
 Loop through x and for each x loop through y
 If we find P(x,y) is true for all x and y, then the statement is true
 If we ever hit a value x for which we hit a value for which P(x,y) is
false, the whole statement is false
 For every x, there exists y: ∀x ∃y P(x, y)
 Loop through x until we find a y that P(x,y) is true
 If for every x, we find such a y, then the statement is true
38

Quantification as loop
 : loop through the values for x until we find an x for
which p(x,y) is always true when we loop through all values for y
 Once found such one x, then it is true
 : loop though the values for x where for each x loop
through the values of y until we find an x for which we find a y
such that p(x,y) is true
 False only if we never hit an x for which we never find y such that p(x,y) is
true
)
,
( y
x
yp
x∀
∃
)
,
( y
x
yp
x∃
∃
39

The order of quantifiers is important unless all the
quantifiers are universal or all the quantifiers are
existential.
 Let Q(x,y) denote “x + y = 0”. What are the truth values of the
quantifications
∀x ∃y Q(x,y)
∃y ∀x Q(x,y)
Order of quantification
40

 Let P(x, y) be the statement “x + y = y + x” and the domain is real
number
∀x∀yP(x, y) ≡ ∀y∀xP(x, y) ?
 The quantification ∀x∀yP(x, y) denotes the proposition “For
all real numbers x, for all real numbers y, x + y = y + x.”
 The statement ∀y∀xP(x, y) says “For all real numbers y, for
all real numbers x, x + y = y + x.”
 Both has the same meaning, and both are true.
41

 Let Q(x, y) denote “x + y = 0.” and the domain is real
number
 What are the truth values of the quantifications
∃y∀xQ(x, y) and ∀x∃yQ(x, y)
 ∃y∀xQ(x, y): “There is a real number y such that for every
real number x, Q(x, y).”
 ∀x∃yQ(x, y): “For every real number x there is a real number
y such that Q(x, y).”
False
True
∃y∀xQ(x, y) ≢ ∀x∃yQ(x, y)
42

Quantifications of Two Variables
43

Quantification with more variables
 ∀x∀y∃zQ(x, y, z): “For all real numbers x and for all real numbers y there is a
real number z such that x + y = z,” is true
 ∃z∀x∀yQ(x, y, z): “There is a real number z such that for all real numbers x
and for all real numbers y it is true that x + y = z,” is false, because there is no
value of z that satisfies the equation x + y = z for all values of x and y.
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44

Translating mathematical statements
 “The sum of two positive integers is always positive”
integers
all
of
consists
bles
both varia
for
domain
the
where
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0
(
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0
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domain
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45

Example
 “Every real number except zero has a multiplicative inverse”
numbers
real
of
consists
bles
both varia
for
domain
the
where
))
1
(
)
0
(( =
∃
→
≠
∀ xy
y
x
x
46
∀x∃y ((x≠0)→ (xy=1))

Translating English Sentences into
Logical Expressions
 “If a person is female and is a parent, then this person is someone’s
mother”
 “For every person x, if person x is female and person x is a parent, then
there exists a person y such that person x is the mother of person y.”
 F(x): “x is female,”
 P(x): “x is a parent,”
 M(x, y): “x is the mother of y.”
∀x((F(x) ∧ P(x)) → ∃yM(x, y))
47

Summary: Predicate Logic
 In this lecture we have learned:
 Predicate logic notation & conventions
 Conversions: predicate logic ↔ clear English
 Meaning of quantifiers, equivalences
 Simple reasoning with quantifiers
48

49

Problem 1
 P(x): “x can speak Russian”
 Q(x): “x knows the computer language C++.”
 Express each of these sentences in terms of P(x), Q(x), quantifiers,
and logical connectives.
 The domain for quantifiers consists of all students at your school.
a) There is a student at your school who can speak Russian and who knows
C++.
∃x(P(x) ^ Q(x))
50

Problem 1
 P(x) be the statement “x can speak Russian”
 Q(x) be the statement “x knows the computer language C++.”
b) There is a student at your school who can speak Russian but who
doesn’t know C++.
c) Every student at your school either can speak Russian or know
C++.
d) No student at your school can speak Russian or know C++.
∃x(P(x) ^ ¬Q(x))
∀x(P(x) ˅ Q(x))
¬ ∃x(P(x) ˅ Q(x)) ∀x ¬(P(x) ˅ Q(x)) ∀x (¬P(x) ^ ¬Q(x))
51

P(x) = “x is a professor” Q(x) = “x is ignorant”
R(x) = “x is vain”
Express the following statements using quantifiers;
logical connectives; and P(x), Q(x), and R(x) where
the universe of discourse is the set of all people.
1. No professors are ignorant
2. All ignorant people are vain
3. No professors are vain
Problem 2
52
∀x(P(x) → ¬Q(x))
∀x(Q(x) → R(x))
∀x(P(x) → ¬ R(x))

Problem 3: Translating Sentences into
Logical Expressions
 Let L(x,y) be the statement “x loves y”, where the domain for both x and y
consists of all people in the world.
1. Everybody loves Jerry.
2. Everybody loves somebody.
3. There is somebody whom everybody loves.
4. Nobody loves everybody.
5. There is somebody whom Lydia does not love.
6. There is somebody whom no one loves.
7. There is exactly one person whom everybody loves.
8. There are exactly two people whom Lynn loves.
9. Everyone loves himself or herself
10. There is someone who loves no one besides himself or herself
53

Programming
Task
Implement using any
programming
language

Programming Task
Let Q(x, y) denote “x + y = 5”
The domain of x and y is the integers {0, 1, 2, 3, 4, 5}.
Implement the code that outputs the truth values of the following
quantifications:
1. ∀x ∀y Q(x, y)
2. ∀x ∃y Q(x, y)
**Display values of variables x and y if the quantification is true.
55

Thinking of Nested
Quantification as Nested Loops
 To see if ∀x ∀y P(x,y) is true,
 loop through the values of x :
 At each step, loop through the values for y.
 If for some pair of x and y, P(x,y) is false, then ∀x ∀y P(x,y) is false and both
the outer and inner loop terminate.
 ∀x ∀y P(x,y) is true if the outer loop ends after stepping through
each x.
56

Thinking of Nested
Quantification as Nested Loops
 To see if ∀x ∃y P(x,y) is true,
 loop through the values of x:
 At each step, loop through the values for y.
 The inner loop ends when a pair x and y is found such that P(x,y) is true.
 If no y is found such that P(x,y) is true the outer loop terminates as ∀x ∃y P(x,y) has
been shown to be false.
 ∀x ∃y P(x,y) is true if the outer loop ends after stepping through each x.
 If the domains of the variables are infinite, then this process can not actually
be carried out.
57

Next Lecture
 Inference Rules
58

Supplementary Material
 1.4.1 Predicate Logic
59

Discrete Mathematics: Lecture 2 - Predicate Logic

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