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 1.1 - Propositional Logic - Exercises - Page 16: 41

Answer

This is a compound proposition made up of two individual propositions; the first proposition is (P v Q v R), and the second is (¬P v ¬Q v ¬R). P v Q v R only evaluates to false when P,Q,R are all FALSE !P v !Q v !R only evaluates to false when are all TRUE So, if P,Q,R are all the same, then one of these propositions will be false. And since we know that TRUE AND FALSE = FALSE, and FALSE AND TRUE = FALSE, then P,Q,R all being the same value will result in the compound proposition being FALSE. If at least one differs from the other, then both propositions will become true, P v Q v R evaluates to true when at least one is true, and !P v !Q v !R evaluates to true when at least one is false

Work Step by Step

Let's use an example, with P = T(rue), Q = F(alse), and R= F(alse). Let's plug them into the compound proposition. (P v Q v R) = (T v F v F) = T. (¬P v ¬Q v ¬R) = (¬T v ¬F v ¬F) = (F v T v T) = T. (P v Q v R) ^ (¬P v ¬Q v ¬R) = (T) ^ (T) = T. Let's use another example using the same truth values, with P= T(rue), Q = T(rue), R = T(rue) (P v Q v R) = (T v T v T) = T. (¬P v ¬Q v ¬R) = (¬T v ¬T v ¬T) = (F v F v F) = F. (P v Q v R) ^ (¬P v ¬Q v ¬R) = (T) ^ (F) = F.
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