CSS342: Quantifiers1 Professor: Munehiro Fukuda

CSS342: Quantifiers2 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: Quantifiers3 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: Quantifiers4 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

CSS342: Quantifiers5 Universally Quantified Statements Example 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: Quantifiers6 Universally Quantified Statements Example 2 For every real number x, x 2 – 1 > 0 is false. To be false, we only need to show a counterexample: x = 1 Proof: If x = 1, the proposition 1 2 – 1 > 0 is false The value 1 is a counterexample. Thus, this universally quantified statement is false

CSS342: Quantifiers7 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 E

CSS342: Quantifiers8 Existentially Quantified Statements Example 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: Quantifiers9 Existentially Quantified Statements Example 2 For some real number x, 1 / (x 2 + 1) > 1 is false. To be false, we need to consider all cases: This in turn means that for all real number x, 1 / (x 2 + 1) ≤ 1 Proof: Since 0 ≤ x 2 for every real number x, 1 ≤ x By dividing both sides of this inequality expression by x / (x 2 + 1) ≤ 1 Is true. Therefore, 1 / (x 2 + 1) >1 is false for every real number x.

CSS342: Quantifiers10 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. AE P(x) = true Domain D ≡ P(x) = true P(x) ≠ true(false)

CSS342: Quantifiers11 xP(x) and xP(x) AE P(x) = false Domain D ≡ P(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.

CSS342: Quantifiers12 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) E A A E

CSS342: Quantifiers13 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 AE

CSS342: Quantifiers14 Revisiting Existentially Quantified Statement Example 2 For some real number x, 1 / (x 2 + 1) > 1 is false. Let P(x) be 1 / (x 2 + 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 / (x 2 + 1) > 1 is false, (i.e., 1 / (x 2 + 1) ≤ 1) Proof: A E Since 0 ≤ x 2 for every real number x, 1 ≤ x By dividing both sides of this inequality expression by x / (x 2 + 1) ≤ 1

CSS342: Quantifiers15 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. AE AE A

CSS342: Quantifiers16 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 A E

CSS342: Quantifiers17 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. A E A AA E E E