1 / 17

# CSS342: Quantifiers - PowerPoint PPT Presentation

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”:

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' CSS342: Quantifiers' - emmett

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### CSS342: Quantifiers

Professor: Munehiro Fukuda

CSS342: Quantifiers

• 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

• 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

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 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 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

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 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 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

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

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

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

• 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

• 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

• 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

• 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

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