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 13: 11

Answer

¬ → ↔ p ∨ q p ∧ q a) p ∧ q b) p ∧ ¬q c) ¬p ∧ ¬q d) p ∨ q e) p → q f) p ⊕ q g) p ↔ q

Work Step by Step

f) It is quite clear from the sentence in English that an exclusive XOR must be employed here. However, if we want to disassemble step by step that sentence, we see that it can be represented as (p ∨ q) ∧ (p → ¬q). For this proposition to be true, one between p and q has to be true (so that the first member is true), but the other ought to be false (otherwise, with both p and q true, the second member will be false). In other words, it is equivalent to the proposition p ⊕ q.
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