# CS 285- Discrete Mathematics

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CS 285- Discrete Mathematics
Lecture 5

Section 1.4 Nested Quantifiers
Nesting of Quantifiers Negating Nested Quantifiers Order of Quantifiers Nested Quantifiers

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

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

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

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

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

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

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

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

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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