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 1.4 - Predicates and Quantifiers - Exercises - Page 56: 54

Answer

$(P(1) \lor P(2) \lor P(3))$ $\land$ $\neg$ $ (P(1) \land P(2))$ $\land$ $\neg$ $ (P(2) \land P(3))$ $\land$ $\neg$ $ (P(1) \land P(3))$

Work Step by Step

$\exists ! x P(x)$ means that there exists exactly one x such that P(x) is true. Domain: 1,2,3 So, we have the following cases: 1) P(1) is True $\quad$ P(2) is false. $\quad$ P(3) is false. 2) P(1) is false $\quad$ P(2) is True. $\quad$ P(3) is false. 3) P(1) is false $\quad$ P(2) is false. $\quad$ P(3) is True. This means that no two of P(1), P(2), P(3) can be simultaneously true. So, the answer is $(P(1) \lor P(2) \lor P(3))$ $\land$ $\neg$ $ (P(1) \land P(2))$ $\land$ $\neg$ $ (P(2) \land P(3))$ $\land$ $\neg$ $ (P(1) \land P(3))$
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