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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)) For all real numbers x, y and z : x+ (y + z) = (x + y) + z) ∀x ∀y (( x > 0) ∧( y < 0) →( xy < 0)) For all real numbers x and y, if x is positive and y is negative, then xy is negative. 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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