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CS 285- Discrete Mathematics Lecture 5

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

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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 3

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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 4

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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 5

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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 6

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

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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 8

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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 9

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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 10

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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 11

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