Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 2 - The Logic of Compound Statements - Exercise Set 2.1 - Page 38: 49

Answer

(a) Commutative law (b) Distributive law (c) Negation law (d) Identity law

Work Step by Step

We use law of logical equivalence to show (p ∨ ∼q) ∧ (∼p ∨ ∼q) ≡ ∼q. Apply commutative law to (p ∨ ∼q) and (∼p ∨ ∼q) to get (p ∨ ∼q) ∧ (∼p ∨ ∼q) ≡ (∼q ∨ p) ∧ (∼q ∨ ∼p). Next apply distributive law to (∼q ∨ p) ∧ (∼q ∨ ∼p) to get (∼q ∨ p) ∧ (∼q ∨ ∼p) ≡ ∼q ∨ (p ∧ ∼p). Then from Negation law we know that (p ∧ ∼p) ≡ c, so ∼q ∨ (p ∧ ∼p) ≡ ∼q ∨ c Finally from Identity law we get ∼q ∨ c ≡ ∼q. Hence (p ∨ ∼q) ∧ (∼p ∨ ∼q) ≡ ∼q using the 4 laws stated above.
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