Explanation Approach: A disjunction is a proposition of the form “ p or q ” and is represented symbolically by p ∨ q . The disjunction p ∨ q is false if both p and q are false; it is true in all other cases. A negation is a proposition of the form “not p ” and is represented symbolically by ∼ p . The proposition ∼ p is true if p is false and vice versa. A conditional statement is a proposition of the form “if p , then q ” and its represented symbolically by p → q . Statements of the form “ p if and only if q ” are called biconditional proposition and are represented symbolically by p ↔ q . A tautology is a statement that is always true. A contradiction is a statement that is always false. Calculation: Construct the truth table for the given expression

11th Edition

ISBN: 9780100478183

Author: Tan

Publisher: YUZU

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Chapter A.4, Problem 9E

To determine

The statement (p→q)↔(∼p∨q) is a tautology, a contradiction, or neither.

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Chapter A Solutions

EBK FINITE MATHEMATICS FOR THE MANAGERI

Ch. A.1 - In Exercises 114, determine whether the statement...Ch. A.1 - Prob. 2E Ch. A.1 - Prob. 3E Ch. A.1 - Prob. 4E Ch. A.1 - Prob. 5E Ch. A.1 - Prob. 6E Ch. A.1 - Prob. 7E Ch. A.1 - Prob. 8E Ch. A.1 - Prob. 9E Ch. A.1 - Prob. 10E

Ch. A.1 - Prob. 11E Ch. A.1 - Prob. 12E Ch. A.1 - Prob. 13E Ch. A.1 - Prob. 14E Ch. A.1 - Prob. 15E Ch. A.1 - Prob. 16E Ch. A.1 - Prob. 17E Ch. A.1 - Prob. 18E Ch. A.1 - Prob. 19E Ch. A.1 - Prob. 20E Ch. A.1 - Prob. 21E Ch. A.1 - Prob. 22E Ch. A.1 - Prob. 23E Ch. A.1 - Prob. 24E Ch. A.1 - Prob. 25E Ch. A.1 - Prob. 26E Ch. A.1 - Prob. 27E Ch. A.1 - Prob. 28E Ch. A.1 - Prob. 29E Ch. A.1 - Let p and q denote the propositions p: The...Ch. A.1 - Prob. 31E Ch. A.1 - Prob. 32E Ch. A.1 - Prob. 33E Ch. A.2 - Prob. 1E Ch. A.2 - Prob. 2E Ch. A.2 - Prob. 3E Ch. A.2 - Prob. 4E Ch. A.2 - In Exercises 1-18, construct a truth table for...Ch. A.2 - Prob. 6E Ch. A.2 - Prob. 7E Ch. A.2 - In Exercises 1-18, construct a truth table for...Ch. A.2 - Prob. 9E Ch. A.2 - Prob. 10E Ch. A.2 - Prob. 11E Ch. A.2 - Prob. 12E Ch. A.2 - Prob. 13E Ch. A.2 - Prob. 14E Ch. A.2 - Prob. 15E Ch. A.2 - Prob. 16E Ch. A.2 - Prob. 17E Ch. A.2 - Prob. 18E Ch. A.2 - If a compound proposition consists of the prime...Ch. A.3 - In Exercises 14, write the converse, the...Ch. A.3 - In Exercises 14, write the converse, the...Ch. A.3 - Prob. 3E Ch. A.3 - Prob. 4E Ch. A.3 - Prob. 5E Ch. A.3 - In Exercises 5 and 6, refer to the following...Ch. A.3 - Prob. 7E Ch. A.3 - Prob. 8E Ch. A.3 - Prob. 9E Ch. A.3 - Prob. 10E Ch. A.3 - Prob. 11E Ch. A.3 - Prob. 12E Ch. A.3 - Prob. 13E Ch. A.3 - Prob. 14E Ch. A.3 - Prob. 15E Ch. A.3 - Prob. 16E Ch. A.3 - Prob. 17E Ch. A.3 - Prob. 18E Ch. A.3 - Prob. 19E Ch. A.3 - Prob. 20E Ch. A.3 - Prob. 21E Ch. A.3 - Prob. 22E Ch. A.3 - Prob. 23E Ch. A.3 - Prob. 24E Ch. A.3 - Prob. 25E Ch. A.3 - Prob. 26E Ch. A.3 - Prob. 27E Ch. A.3 - Prob. 28E Ch. A.3 - Prob. 29E Ch. A.3 - Prob. 30E Ch. A.3 - Prob. 31E Ch. A.3 - Prob. 32E Ch. A.3 - Prob. 33E Ch. A.3 - Prob. 34E Ch. A.3 - Prob. 35E Ch. A.3 - Prob. 36E Ch. A.3 - Prob. 37E Ch. A.3 - Prob. 38E Ch. A.4 - Prove the idempotent law for conjunction, ppp.Ch. A.4 - Prob. 2E Ch. A.4 - Prove the associative law for conjunction,...Ch. A.4 - Prob. 4E Ch. A.4 - Prove the commutative law for conjunction, pqqp.Ch. A.4 - Prob. 6E Ch. A.4 - Prob. 7E Ch. A.4 - Prob. 8E Ch. A.4 - Prob. 9E Ch. A.4 - Prob. 10E Ch. A.4 - Prob. 11E Ch. A.4 - Prob. 12E Ch. A.4 - Prob. 13E Ch. A.4 - Prob. 14E Ch. A.4 - Prob. 15E Ch. A.4 - Prob. 16E Ch. A.4 - Prob. 17E Ch. A.4 - In exercises 9-18, determine whether the statement...Ch. A.4 - Prob. 19E Ch. A.4 - Prob. 20E Ch. A.4 - In Exercises 21-26, use the laws of logic to prove...Ch. A.4 - Prob. 22E Ch. A.4 - In Exercises 21-26, use the laws of logic to prove...Ch. A.4 - In Exercises 21-26, use the laws of logic to prove...Ch. A.4 - In Exercises 21-26, use the laws of logic to prove...Ch. A.4 - Prob. 26E Ch. A.5 - Prob. 1E Ch. A.5 - Prob. 2E Ch. A.5 - Prob. 3E Ch. A.5 - Prob. 4E Ch. A.5 - Prob. 5E Ch. A.5 - Prob. 6E Ch. A.5 - Prob. 7E Ch. A.5 - Prob. 8E Ch. A.5 - Prob. 9E Ch. A.5 - In Exercises 116, determine whether the argument...Ch. A.5 - Prob. 11E Ch. A.5 - Prob. 12E Ch. A.5 - Prob. 13E Ch. A.5 - Prob. 14E Ch. A.5 - Prob. 15E Ch. A.5 - Prob. 16E Ch. A.5 - In Exercises 17-22, represent the argument...Ch. A.5 - Prob. 18E Ch. A.5 - In Exercises 17-22, represent the argument...Ch. A.5 - In Exercises 17-22, represent the argument...Ch. A.5 - In Exercises 17-22, represent the argument...Ch. A.5 - Prob. 22E Ch. A.5 - Prob. 23E Ch. A.5 - Prob. 24E Ch. A.5 - Prob. 25E Ch. A.6 - In Exercises 1-5, find a logic statement...Ch. A.6 - Prob. 2E Ch. A.6 - Prob. 3E Ch. A.6 - Prob. 4E Ch. A.6 - Prob. 5E Ch. A.6 - Prob. 6E Ch. A.6 - Prob. 7E Ch. A.6 - Prob. 8E Ch. A.6 - Prob. 9E Ch. A.6 - Prob. 10E Ch. A.6 - Prob. 11E Ch. A.6 - Prob. 12E Ch. A.6 - Prob. 13E Ch. A.6 - In Exercise 12-15, find a logic statement...Ch. A.6 - Prob. 15E Ch. A.6 - Prob. 16E

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The statement ( p → q ) ↔ ( ∼ p ∨ q ) is a tautology, a contradiction, or neither.

Solution Summary: The author explains that the statement (pto q) is a tautology.

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The statement (p→q)↔(∼p∨q) is a tautology, a contradiction, or neither.

Chapter A Solutions