Discrete Mathematics and Its Applications, Seventh Edition

Published by McGraw-Hill Education
ISBN 10: 0073383090
ISBN 13: 978-0-07338-309-5

Chapter 1 - Section 1.3 - Propositional Equivalences - Exercises - Page 35: 30

Answer

$ (p∨q)∧(¬p∨r)→(q∨r)$ is a tautology

Work Step by Step

We have, $ [(p∨q)∧(¬p∨r)]→(q∨r) ≡ ¬[(p∨q)∧(¬p∨r)]∨(q∨r) ≡[¬(p∨q)∨¬(¬p∨r)]∨(q∨r) ≡[(¬p∧¬q)∨(¬(¬p)∧¬r)]∨(q∨r) ≡[(¬p∧¬q)∨(p∧¬r)]∨(q∨r) ≡[((¬p∧¬q)∨p)∧((¬p∧¬q)∨¬r)]∨(q∨r) ≡[((¬p∨p)∧(¬q∨p))∧((¬p∨¬r)∧(¬q∨¬r))]∨(q∨r) ≡[(T∧(¬q∨p))∧((¬p∨¬r)∧(¬q∨¬r))]∨(q∨r) ≡[(¬q∨p)∧((¬p∨¬r)∧(¬q∨¬r))]∨(q∨r) ≡([(¬q∨p)∧((¬p∨¬r)∧(¬q∨¬r))]∨q)∨r ≡[((¬q∨p)∨q)∧(((¬p∨¬r)∨q)∧((¬q∨¬r)∨q))]∨r ≡[((¬q∨q)∨p)∧(((¬p∨¬r)∨q)∧((¬q∨q)∨¬r))]∨r ≡[(T∨p)∧(((¬p∨¬r)∨q)∧(T∨¬r))]∨r ≡[T∧(((¬p∨¬r)∨q)∧T)]∨r ≡[T∧((¬p∨¬r)∨q)]∨r ≡((¬p∨¬r)∨q)∨r ≡(¬p∨¬r)∨(q∨r) ≡(¬p∨¬r)∨(r∨q) ≡((¬p∨¬r)∨r)∨q ≡(¬p∨(¬r∨r))∨q ≡(¬p∨T)∨q ≡T∨q ≡T $
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