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CS 285- Discrete Mathematics

CS 285- Discrete Mathematics. Lecture 5. Section 1.4 Nested Quantifiers. Nesting of Quantifiers Negating Nested Quantifiers Order of Quantifiers. Nesting of Quantifiers. Nested Quantifiers are quantifiers that occur within the scope of other quantifiers Example:

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CS 285- Discrete Mathematics

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  1. CS 285- Discrete Mathematics Lecture 5

  2. Section 1.4 Nested Quantifiers • Nesting of Quantifiers • Negating Nested Quantifiers • Order of Quantifiers Nested Quantifiers

  3. Nesting of Quantifiers • Nested Quantifiers are quantifiers that occur within the scope of other quantifiers • Example: • P(x,y) = x likes y, where the u.d. for x & y is all people ( a predicate with 2 free variables (f.v.) • ∃yP(x,y) = There is someone whom x likes ( a predicate with 1 f.v.) • ∀x (∃y P(x,y)) = Everyone has someone whom they like. ( a PROPOSITION) Nested Quantifiers

  4. Translating Statements into Nested Quantifiers • Translate the following statements, where the u.d consists of all real numbers: • ∀x ∀y (x+y = y+x) For all real numbers x and y : x + y = y+x • ∀x ∃ y (x+y = 0) For every real number x there is a real number y such that x+ y =0 • ∀x ∀y ∀z (x+ (y + z) = (x + y) + z)) Forall real numbers x, y and z : x+ (y + z) = (x + y) + z) • ∀x ∀y (( x > 0) ∧( y < 0) →( xy < 0)) Forall real numbers x and y, if x is positive and y isnegative, thenxyisnegative. Nested Quantifiers

  5. Examples • If R(x,y)=“x relies upon y,” express the following in unambiguous English: • ∀x(∃y R(x,y))= Everyone has someone to rely on. • ∃y(∀x R(x,y))= There’s an overburdened soul whom everyone relies upon (including himself) • ∃x(∀y R(x,y))= There’s some needy person who relies upon everybody (including himself) • ∀y(∃x R(x,y))= Everyone has someone who relies upon them. • ∀x(∀y R(x,y))= Everyone relies upon everybody, (including themselves)! Nested Quantifiers

  6. Natural Language Ambiguity 1. “Everybody likes somebody.” • For everybody, there is somebody they like, • ∀x ∃y Likes(x,y) • or, there is somebody (a popular person) whom everyone likes? • ∃y ∀x Likes(x,y) 2. “Somebody likes everybody.” • Same problem: Depends on context, emphasis. Nested Quantifiers

  7. Negating Nested Quantifiers • By using Demorgan’s equivalence laws: • ¬ ∀x P(x) ⇔ ∃x ¬P(x) • ¬ ∃x P(x) ⇔ ∀x ¬P(x) Nested Quantifiers

  8. Equivalence laws and Conventions • ∀x ∀y P(x,y) ⇔ ∀y ∀x P(x,y) ∃x ∃y P(x,y) ⇔ ∃y ∃x P(x,y) • ∀x (P(x) ∧ Q(x)) ⇔ (∀x P(x)) ∧ (∀x Q(x)) • ∃x (P(x) ∨ Q(x)) ⇔ (∃x P(x)) ∨ (∃x Q(x)) • parenthesize ∀x (P(x) ∧ Q(x)) • Consecutive quantifiers of the same type can be combined: ∀x ∀y ∀z P(x,y,z) ⇔ ∀x,y,z P(x,y,z) or even ∀xyz P(x,y,z) • All quantified expressions can be reduced to the canonical alternating form : ∀x1∃x2∀x3∃x4…P(x1, x2, x3, x4,…) Nested Quantifiers

  9. Order of Quantifiers • The order of quantifiers is important when translating any statement unless they are all universal quantifiers or existential quantifiers. • ∀x ∀yP(x, y) ⇔ ∀y ∀xP( x, y)? YES! • ∀x∃ yP(x, y) ⇔ ∃ y ∀xP( x, y)? NO! Different Meaning !!! • ∀x[P(x) ∧ Q(x)] ⇔ ∀x P( x) ∧ ∀x Q( x)? YES! • ∀x[P(x)→Q(x)] ⇔ ∀x P( x) →∀x Q( x)? NO! Nested Quantifiers

  10. Exercise --- I • Translate the following statements into logical ones: • There is a women who has taken a flight on every airline in the world. (u.d. all women in the world) ∃w∀ a ∃f (P(w,f) ∧ Q(f,a)) • There doesn’t exist a women who has taken a flight on every airline in the world. ¬∃w∀ a ∃f (P(w,f) ∧ Q(f,a)) ⇔ ∀w∃ a∀ f (¬ P(w,f) ∨ ¬ Q(f,a)) Nested Quantifiers

  11. Exercise --- II • “There is no store that has no students who shop there.” • S(x, y): “x shops in y” • T (x): “x is a student” • where the universe for x consists of people and the universe for y consists of stores: • Rewriting the above statement: All stores have students who shop in them. Thus if you are a student, then you shop in one of the stores. • we have ¬∃y ∀x (T (x)→¬S(x, y)). Nested Quantifiers

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