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: 51

Answer

Let the domain consist of y and z for which P(y) is false Q(y) is true P(z) is true Q(z) is false. $\exists xP(x) \land \exists x Q(x)$ is then true (because $\exists xP(x)$ and $\exists xQ(x)$ is true) and $\exists x(P(x) \land Q(x))$ ls false (since $P(x)$ and $Q(x)$ are never true at the same time). Since the two expressions do not have the same truth value for this case, the two expressions can’t be logically equivalent.

Work Step by Step

See answer
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.