Explanation Given: For all integers a , b and c if a divides b and a not divides c then a not divides ( b + c ) . Calculation: Suppose not, that is, suppose there exist integers a , b and c such that: a/b by definition of divisibility, there exists integers r and s such that b = a r and b + c = a s by substitution, a r + c = a s subtracting a r from both sides gives c = a s − a r = a ( s − r ) but s − r is an integer because r and s are integers. Hence, by definition of divisibility, etc. this contradiction our supposition that a , b hence, supposition is false and statement is true. By De Morgan’s law it has show that for all integers a , b and c : a / ( b + c ) → ∼ ( a / b ) ∨ ( a / c ) But by logical equivalence of p ∧ ∼ q → r and p ∧ ∼ r → q

11th Edition

ISBN: 9781133417071

Author: EPP

Publisher: CENGAGE LEARNING - CONSIGNMENT

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Chapter 4.5, Problem 29ES

To determine

Toprove: p→q∨r, is suffices to prove either p∧∼q→r or p∧∼r→q .

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Chapter 4.5, Problem 28ES

Chapter 4.5, Problem 30ES

Chapter 4 Solutions

EBK DISCRETE MATHEMATICS: INTRODUCTION

Ch. 4.1 - Prob. 1ES Ch. 4.1 - Prob. 2ES Ch. 4.1 - Prob. 3ES Ch. 4.1 - Prob. 4ES Ch. 4.1 - Prob. 5ES Ch. 4.1 - Prob. 6ES Ch. 4.1 - Prob. 7ES Ch. 4.1 - Prob. 8ES Ch. 4.1 - Prob. 9ES Ch. 4.1 - Prob. 10ES

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Toprove : p → q ∨ r , is suffices to prove either p ∧ ∼ q → r or p ∧ ∼ r → q .

Solution Summary: The author explains that pto qvee r, is suffices to prove either

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Toprove: p→q∨r, is suffices to prove either p∧∼q→r or p∧∼r→q .

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Chapter 4 Solutions