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 2.2 - Page 50: 47

Answer

a. ~(p∧∼q)∨r b. ~((p∧∼q)∧~r)

Work Step by Step

a. p∧∼q→r =~(p∧∼q)∨r Substitute using the equivalence p → q =∼p ∨ q, where p = (p∧∼q), and q = r b. Then using the new form: ~(p∧∼q)∨r =~(~(~(p∧∼q)))∧~r) Substitute using the equivalence p ∨ q =∼(∼p∧ ∼q), where p = ~(p∧∼q) and q= r =~((p∧∼q)∧~r) Simplify double inversions
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