a1-sol.pdf - CSI 2101 Discrete Structures Prof Lucia Moura Winter 2012 University of Ottawa Homework Assignment#1(100 points weight 5 Due Thursday Feb 9

# a1-sol.pdf - CSI 2101 Discrete Structures Prof Lucia Moura...

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CSI 2101 Discrete Structures Winter 2012 Prof. Lucia Moura University of Ottawa Homework Assignment #1 (100 points, weight 5%) Due: Thursday Feb 9, at 1:00 p.m. (in lecture); assignments with lateness between 1min-24hs will have a discount of 10%; after 24hs, not accepted; please drop off late assignments under my office door (STE5027). Propositional Logic 1. (12 points) Use logical equivalences to show that [( p q ) ( p r ) ( q r )] r is a tautology. It is sufficient to show that if we assume the premise ( p q ) ( p r ) ( q r ), then we can derive the conclusion r . Here is such an argument: 1. p q Simplification 2. p r ≡ ¬ p r Simplification 3. q r ≡ ¬ q r Simplification 4. q r Resolution (1) and (2) 5. r r r Resolution (3) and (4) Thus, the statement is a tautology. 2. (12 points; each 2+2+2 points=truth table+DNF+CNF) For each of the following compound propositions give its truth table and derive an equivalent compound proposition in disjunctive normal formal (DNF) and in conjunc- tive normal form (CNF). (a) ( p q ) r We write out the truth table: p q r p q ( p q ) r T T T T T T T F T F T F T F T T F F F T F T T T T F T F T F F F T T T F F F T F 1 Subscribe to view the full document.

To get disjunctive normal form, we take the conjunction of the values for p , q , and r for each row where the final evaluation is T, and then take their disjunction: ( p q r ) ( p ∧ ¬ q r ) ( p ∧ ¬ q ∧ ¬ r ) ( ¬ p q r ) ( ¬ p ∧ ¬ q r ) To get conjunctive normal form, we take the conjunction of the values for p , q , and r for each row where the final evaluation is F, and then take their disjunction: ( p q ∧ ¬ r ) ( ¬ p q ∧ ¬ r ) ( ¬ p ∧ ¬ q ∧ ¬ r ) Now we take the negation of this entire expression, and repeatedly using DeMor- gan’s, we derive: ( ¬ p ∨ ¬ q r ) ( p ∨ ¬ q r ) ( p q r ) (b) ( p ∧ ¬ q ) ( p r ) We write out the truth table: p q r p ∧ ¬ qq p r ( p ∧ ¬ q ) ( p r ) T T T F T T T T F F F F T F T T T T T F F T F T F T T F F F F T F F T T F F T F F F F F F F T T To get disjunctive normal form, we take the conjunction of the values for p , q , and r for each row where the final evaluation is T, and then take their disjunction: ( p q r ) ( p ∧ ¬ q r ) ( p ∧ ¬ q ∧ ¬ r ) ( ¬ p q ∧ ¬ r ) ( ¬ p ∧ ¬ q ∧ ¬ r ) To get conjunctive normal form, we take the conjunction of the values for p , q , and r for each row where the final evaluation is F, and then take their disjunction: ( p q ∧ ¬ r ) ( ¬ p q r ) ( ¬ p ∧ ¬ q r ) Now we take the negation of this entire expression, and repeatedly using DeMor- gan’s, we derive: ( ¬ p ∨ ¬ q r ) ( p ∨ ¬ q ∨ ¬ r ) ( p q ∨ ¬ r ) 2 Predicate Logic 3. (15 points) For each of the given statements: 1 - Express each of the statements using quantifiers and propositional functions. Subscribe to view the full document. • Winter '11
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