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Unformatted text preview: MACM 101 Discrete Mathematics I Outline Solutions to Exercises on Propositional Logic 1. Construct a truth table for the following compound proposition: ( p q ) ( q r ) . p q r ( p q ) ( q r ) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2. Show that the following compound statement is a contradiction (( p q ) ( p r )) ( p ( q r )) . Method 1. Construct a truth table. Method 2. If this statement is not a contradiction, it is true for some truth values of p , q , and r . We try to find these values. Since the statement is an exclusive OR, it is true only if one part is true and the other is false. Suppose (( p q ) ( p r )) is false. It is a disjunction, therefore both part of it must be false. If p q is false, p must be true, q must be true. Similarly, if p r is false, p must be true, r must be true. It is easy to check that with this truth values the second part, p ( q r ) is also false, therefore the whole statement is false. The case when p ( q r ) is false is similar. 3. Show that ( p q ) and ( p q ) are logically equivalent. Method 1. Construct the truth tables for both statements and compare. Method 2. Use logical equivalences. ( p q ) ( p q ) ( q p ) expression for biconditional ( p q ) ( q p ) expression for implication, twice ( q p ) ( p q ) Commutative Law ( ( q ) p ) (( ( p ) q ) double negation (( q ) p ) (( p ) q ) expression for implication ( q p ) expression for biconditional 4. Show that ( p q ) r and ( p r ) ( q r ) are not logically equivalent. Do not use truth tables....
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- Summer '08