SolHW1f07

# SolHW1f07 - HW1S07 CS 336 1 Determine the truth value of...

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Unformatted text preview: HW1S07 CS 336 1. Determine the truth value of the following statements a. ( ∀ x| x ∈Ζ : ( ∃ y| y ∈Ζ : x 2 <y))=T b. ( ∃ y| y ∈Ζ : ( ∀ x| x ∈Ζ : x 2 <y))= F c. What you notice from parts a. and b.? The answers are different. What are the implications? Order is important for nested quantifiers. 2. Rewrite these with equivalent statements with the negation only within predicates. a. ¬ ( ∀ x| x ∈Ζ : ( ∃ y| y ∈Ζ : x 2 <y)) = ( ∃ x| x ∈ Z: ( ∀ y| y ∈ Z: x ≥ y)) b. ¬ ( ∃ y| y ∈Ζ : ( ∀ x| x ∈Ζ : x 2 <y)) = ( ∀ y| y ∈ Z: ( ∃ x| x ∈ Z: x ≥ !y)) c. How might you check if your statements are equivalent? Check the truth value of each. Are they? Yes. 3. Prove the following a. a. (b^(b → c)) → c Proof: (b^(b → c)) → c ↔ < → > (b ∧ ( ¬ b ∨ c) ) → c ↔ < → > ¬ (b ∧ ( ¬ b ∨ c) ) ∨ c ↔ < Distributivity > ¬ ((b ∧ ¬ b) ∨ (b ∧ c) ) ∨ c ↔ < Contradiction> ¬ ( F ∨ (b ∧ c) ) ∨ c ↔ < Commutativity; ∨-simplification > ¬ (b ∧ c) ∨ c ↔ < DeMorgan’s Law > ( ¬ b ∨ ¬ c) ∨ c ↔ < Associativity > ¬ b ∨ ( ¬ c ∨ c) ↔ < Commutativity; Excluded Middle > ¬ b ∨ T ↔ < ∨-simplification > T QED b. p ∧ q → p Proof: p ∧ q → p ↔ < → > ¬ (p ∧ q) ∨ p ↔ < DeMorgan’s Law > ( ¬ p ∨ ¬ q) ∨ p ↔ < Commutativity > p ∨ (¬ p ∨ ¬ q) ↔ < Associativity > (p ∨ ¬ p) ∨ ¬ q ↔ < Excluded Middle > T ∨ ¬ q ↔ < Commutativity; ∨-simplification >...
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## This note was uploaded on 09/20/2008 for the course CS 336 taught by Professor Myers during the Spring '08 term at University of Texas.

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SolHW1f07 - HW1S07 CS 336 1 Determine the truth value of...

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