CS 285- Discrete Mathematics

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