CSS342: Quantifiers

CSS342: Quantifiers 1

CSS342: Quantifiers

Professor: Munehiro Fukuda


Review of Propositions• Proposition: a statement that is either true or false, but not

both• Example:

– 1 < 4 is true.– 2 > 5 is false.– 3 is an odd number– Then, how about x is an odd number?

• The statement “X is an odd number”:– is true if x = 103– is false if x = 8– Most of the statements in math and CS use variables.

• We need to extend the system of logic!


Propositional Functions• P(x): a statement involving the variable x

– Example: x is an odd number.– P(x) itself is not a proposition.– For each x in the domain D of discourse of P, if D is the set of

positive integers, P(x) is a proposition• P(1): 1 is an odd number (= true).• P(2): 2 is an odd number (= false).• …• Either true or false

• X: a free variable, (i.e., free to roam over the domain D) • How about for every x or for some x, P(x) is true/false?

– Most statements in math and CS use such phrases.


Universally Quantified Statements

• x, P(x)– Meaning: for every x, P(x), for all x, P(x), or

for any x, P(x)– : a universal quantifier– P(x): a universally qualified statement– X: a bound variable, (i.e., bound by the

quantifier )– is true: if P(x) is true for every x in D– is false: if P(x) is false for at least one x in D

A

A

A


Universally Quantified StatementsExample 1

• For every real number x, if x > 1, then x + 1 > 1is true.

• To be true, we need to consider all cases:x ≤ 1 and x > 1

• Proof:(1 ) If x ≤ 1,

the hypothesis x > 1 is false,

thus the conditional proposition is true.

(2) If x > 1,

x + 1 > x (always true)

since x > 1, x + 1 > x > 1

Thus, the conditional proposition is true.

Therefore, this universally quantified statement is true.


Universally Quantified StatementsExample 2

• For every real number x, x2 – 1 > 0is false.

• To be false, we only need to show a counterexample:x = 1

• Proof:

If x = 1, the proposition 12 – 1 > 0 is false

The value 1 is a counterexample.

Thus, this universally quantified statement is false


Existentially Quantified Statements

• x, P(x)– Meaning: for some x, P(x), for at least one x, P(x),

or there exists x such that, P(x)– : an existential quantifier– P(x): a existentially qualified statement– X: a bound variable, (i.e., bound by the

quantifier )– is true: if P(x) is true for at least one x in D– is false: if P(x) is false for every x in D

E

E

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Existentially Quantified StatementsExample 1

• For some positive integer n, if n is prime, then n + 1, n + 2, n + 3, and n + 4 are not prime.is true.

• To be true, we only need to show at least one case makes it true:x = 1

• Proof: If n = 23, n + 1 = 24 (=3 * 8) is not prime

n + 2 = 25 (=5 * 5) is not prime

n + 3 = 26 (=2 * 13) is not prime

n + 4 = 27 (=3 * 9) is not prime

Thus, the proposition is true.

Therefor, this existentially quantified statement is true.


Existentially Quantified StatementsExample 2

• For some real number x, 1 / (x2 + 1) > 1is false.

• To be false, we need to consider all cases:This in turn means that for all real number x, 1 / (x2 + 1) ≤ 1

• Proof:Since 0 ≤ x2 for every real number x,

1 ≤ x2 + 1

By dividing both sides of this inequality expression by

x2 + 1

1 / (x2 + 1) ≤ 1

Is true.

Therefore, 1 / (x2 + 1) >1 is false for every real number x.


xP(x) and xP(x)

• If it is not the case that P(x) is true for every x, there is a counterexample showing that P(x) is not true for at least one x.

A E

P(x) = true

P(x) = trueP(x) = true

Domain D

≡

Domain D

P(x) = trueP(x) = true

P(x) ≠ true(false)


xP(x) and xP(x)

AE

P(x) = false

P(x) = falseP(x) = false

Domain D

≡

Domain D

P(x) = falseP(x) = false

P(x) = true

• If it is not the case that P(x) is true for some x, P(x) is false for every x.


Generalized De Morgan Laws for Logic

If P is a propositional funciton, each pair of propositions in (a) and (b) is logically equivalent, (i.e., has the same truth values.)

A) xP(x) and xP(x)

B) xP(x) and xP(x)

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A

A

E


Generalized De Morgan Laws for Logic (Cont’d)

• For every x, P(x) means– P(1) && P(2) && … && P(n)

• For some x, P(x) means– P(1) || P(2) || … || P(n)

• x, P(x) ≡ x, P(x) means– !(P(1) && P(2) && … && P(n)) ≡ !P(1) || !P(2) || … || !P(n)

• x, P(x) ≡ x, P(x) means– !(P(1) || P(2) || … || P(n)) ≡ !P(1) && !P(2) && … && !P(n))

AE

A E


Revisiting Existentially Quantified Statement Example 2

• For some real number x, 1 / (x2 + 1) > 1 is false.• Let P(x) be 1 / (x2 + 1) > 1, then

x, P(x)• According to De Morgan Laws,

we may prove x, P(x)This in turn means that for all real number x,1 / (x2 + 1) > 1 is false, (i.e., 1 / (x2 + 1) ≤ 1)

• Proof:

A

ESince 0 ≤ x2 for every real number x, 1 ≤ x2 + 1

By dividing both sides of this inequality expression by x2 + 1

1 / (x2 + 1) ≤ 1


Two-Variable Propositional Function

• P(x, y): x + y = 0• x, y, x + y = 0 is true.

– Proof:for any x, we can find at least y = -x.Thus, x + y = x – x = 0

• y, x, x + y = 0 is false.– Proof: if it is true, the statement can be replaced

with constant Y x, x + Y = 0However, we can choose x = 1 – Y, such

thatx + Y = 1 – Y + Y = 1 ≠ 0.

A E

AE

A


The Logic Game

• Given a propositional function, P(x, y)You and your opponent play a logic game.– Your goal: make P(x, y) true– You can: choose a bound variable of to make it true– Your opponent, Farley’s goal: make P(x, y) false– Farley can: choose a bound variable of to make it false

• Play with P(x, y): x + y = 0

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E


The Logic Game (Cont’d)• x, y, P(x, y):

– No matter what Farley chooses for x, you choose y = -x.– You win. P(x, y) is true.

• x, y, P(x, y):– Regardless of what you choose for x, Farley chooses y = 1- x.– Farley wins. P(x, y) is false.

• x, y, P(x, y):– Farley chooses x and y such that x + y ≠ 0, (e.g., x = 0, y = 1)– Farley wins. P(x, y) is false.

• x, y, P(x, y)– You choose x and y such that x + y = 0, (e.g., y = -x or x=1, y= -1)– You win. P(x, y) is true.

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CSS342: Quantifiers