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CSS342: Quantifiers

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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

- 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

- 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

- 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

- 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

Domain D

P(x) ≠ true(false)

P(x) = true

P(x) = true

≡

P(x) = true

P(x) = true

P(x) = true

A

E

- 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

E

A

- 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

- Proof:for any x, we can find at least y = -x.
- 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.

- Proof: if it is true, the statement can be replaced with constant Y

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