13+PS1+Solutions

13+PS1+Solutions - Handout #13 April 11, 2011 CS103 Robert...

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Handout #13 CS103 April 11, 2011 Robert Plummer Problem Set #1 -- Solutions 1. Truth Tables (a) (b) (c) (d) 7 1 5 4 2 6 3 3 1 4 2 4 1 5 3 2 2 1 5 4 3
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2 (e) P Q R T T T T T T F T T F T F T F F F F T T T F T F F F F T T F F F F (f) (P Q R) (P Q ¬R) (¬P Q R) (¬P ¬Q R) (P Q) (¬P R) 2. Equivalences (a) The expressions are not equivalent because the columns under the main connectives are not identical. (b) The expressions are equivalent because the columns under the main connectives are identical. These are the same as Q. These are the same as R.
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3 (c) a. Exercise 49 states that p q ¬(p q). So p p ¬(p p) and thus p p ¬p by the Idempotent Law. b. Again using Exercise 49, we have ( p q) (p q) ¬( ¬(p q) ¬(p q) ) ( p q) (p q) ¬( ¬(p q) ) by the Idempotent Law. ( p q) (p q) (p q) by Double Negation c. The derivations above show that negation and disjunction can be expressed in terms of . Exercise 45 shows that negation and disjunction form a functionally complete set, so is functionally complete by itself. 3. Tautologies (a) No, because not all T's appear below the main connective in the truth table.
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This note was uploaded on 06/01/2011 for the course EE 103 taught by Professor Plummer during the Spring '11 term at Stanford.

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13+PS1+Solutions - Handout #13 April 11, 2011 CS103 Robert...

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