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Understanding Quantifiers in Propositional Logic: Examples & Applications

Explore the concepts of quantifiers in propositional logic with examples and applications in math and computer science. Learn about universally quantified statements, existentially quantified statements, domain, and generalized De Morgan laws.

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Understanding Quantifiers in Propositional Logic: Examples & Applications

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  1. CSS342: Quantifiers Professor: Munehiro Fukuda CSS342: Quantifiers

  2. 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

  3. 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

  4. 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

  5. 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

  6. 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

  7. 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

  8. 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

  9. 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

  10. 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

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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

  16. 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

  17. 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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