Explanation Given: p ∨ _ q ≡ ∼ [ ( p ∨ q ) → ( p ∧ q ) ] . Explanation: Since the number of component statements is 2 in the compound statement, the number of rows must be 2 2 = 4. First of all, combinations of the truth values of p and q are written. Let’s take the left-hand side. The compound statement with ∨ _ operator between p and q is written where true value occurs when either of the component statements p or q is true but not both. Let’s take the right-hand side. The compound statement with o r operator between p and q is written where true value occurs when either of the component statements p or q is true. Then, the compound statement with a n d operator between p and q is written where true value occurs when both of the component statements p and q are true. Then, the compound statement is the → operator between ( p ∨ q ) and ( p ∧ q ) where false value occurs only when ( p ∨ q ) is true and ( p ∨ q ) is false

11th Edition

ISBN: 9780321979438

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 6, Problem 40RE

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To construct: A truth table to show that p∨_q≡∼[(p∨q)→(p∧q)].

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Chapter 6, Problem 39RE

Chapter 6, Problem 41RE

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Finite Mathematics (11th Edition)

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To construct: A truth table to show that p ∨ _ q ≡ ∼ [ ( p ∨ q ) → ( p ∧ q ) ] .

Solution Summary: The author explains that the truth table for the provided compound statement is as follows.

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To construct: A truth table to show that p∨_q≡∼[(p∨q)→(p∧q)].

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Chapter 6 Solutions