CSE 20 Quiz 1 Solutions

CSE 20 Quiz 1 Solutions - b ∃ x in R such that(x 2 2x...

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Answers to Quiz #1 1. _p_|_q_|_r_|_(p q)___ ___(~p r)__||__ __||__(q r)_ T | T | T | T T T || T || T T | T | F | T F F || T || T T | F | T | T T T || T || T 6 pts T | F | F | T F F || T || F F | T | T | T T T || T || T F | T | F | T T T || T || T F | F | T | F F T || T || T F | F | F | F F T || T || F 2. p (p q) p ( ~p q) by result shown in class (p ~p) (p q) by the distributive law c (p q) by the negation law (p q) c by the commutative law (p q) by the identity law (p q) since they have different truth tables. 3. a. integer x 12 | x and x is not divisible by 4 or x is not divisible by 3.
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Unformatted text preview: b. ∃ x in R such that (x 2 + 2x + 1( < 0 and x = -1 c. ∀ computer science student x, is is not a math major. 4. Let R = rain F = foggy S = Sailing Race L = Lifesaving Demo T = Trophy The argument can be written as (~R ∨ ~F) → (S ∧ L) S → T ~T ∴ R ~T ∴→ ~S by Modus Tollens ~S → ~(S ∧ L) by definition of the conjunction ~(S ∧ L) → ~(~R ∨ ~F) by Modus Tollens ~(~R ∨ ~F) ≡ (R ∧ F) by DeMorgan’s Laws (R ∧ F) → R by Specialization Therefore, the argument is valid. 5. a. ~s → ~r b. s → r c. ~r → ~s 6. b and c are true. 7. a. 1589 b. 14eb c. 111010101101001111 8. 01110001...
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CSE 20 Quiz 1 Solutions - b ∃ x in R such that(x 2 2x...

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