Css342 quantifiers
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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”:

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CSS342: Quantifiers

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Css342 quantifiers

CSS342: Quantifiers

Professor: Munehiro Fukuda

CSS342: Quantifiers


Review of propositions

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!

CSS342: Quantifiers


Propositional functions

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.

CSS342: Quantifiers


Universally quantified statements

Universally Quantified Statements

A

  • 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

CSS342: Quantifiers


Universally quantified statements example 1

Universally Quantified StatementsExample 1

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

    is 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.

CSS342: Quantifiers


Universally quantified statements example 2

Universally Quantified StatementsExample 2

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

    is 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

CSS342: Quantifiers


Existentially quantified statements

Existentially Quantified Statements

E

  • 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

CSS342: Quantifiers


Existentially quantified statements example 1

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.

CSS342: Quantifiers


Existentially quantified statements example 2

Existentially Quantified StatementsExample 2

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

    is 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.

CSS342: Quantifiers


Xp x and xp x

Domain D

Domain D

P(x) ≠ true(false)

P(x) = true

P(x) = true

P(x) = true

P(x) = true

P(x) = true

A

E

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.

CSS342: Quantifiers


Xp x and xp x1

E

A

xP(x) and xP(x)

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

Domain D

Domain D

P(x) = true

P(x) = false

P(x) = false

P(x) = false

P(x) = false

P(x) = false

CSS342: Quantifiers


Generalized de morgan laws for logic

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.)

  • xP(x)and xP(x)

  • xP(x)and xP(x)

A

E

E

A

CSS342: Quantifiers


Generalized de morgan laws for logic cont d

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))

A

E

E

A

CSS342: Quantifiers


Revisiting existentially quantified statement example 2

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:

E

A

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

CSS342: Quantifiers


Two variable propositional function

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 = 0

      However, we can choose x = 1 – Y, such that

      x + Y = 1 – Y + Y = 1 ≠ 0.

A

E

E

A

A

CSS342: Quantifiers


The logic game

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

E

A

CSS342: Quantifiers


The logic game cont d

The Logic Game (Cont’d)

A

E

  • 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.

E

A

A

A

E

E

CSS342: Quantifiers


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