Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 6 - Set Theory - Exercise Set 6.3 - Page 373: 31

Answer

$For\,\,all\,\,sets\,\,A,\,B,\,and\,C,\\ A \cup (B - A) = A \cup B.\\ Proof:\\ A \cup (B - A)=A \cup (B \cap A^c)\\ by(set\,\,dif\! ference\,\,law )\\ =(A\cup B)\cap (A\cup A^c)\\ by(distributive\,\,law)\\ =(A\cup B)\cap U\\ by(complement\,\,law\,for\,\cup )\\ =A\cup B \\ by(identity\,law\,for\,\cap )\\$

Work Step by Step

$For\,\,all\,\,sets\,\,A,\,B,\,and\,C,\\ A \cup (B - A) = A \cup B.\\ Proof:\\ A \cup (B - A)=A \cup (B \cap A^c)\\ by(set\,\,dif\! ference\,\,law )\\ =(A\cup B)\cap (A\cup A^c)\\ by(distributive\,\,law)\\ =(A\cup B)\cap U\\ by(complement\,\,law\,for\,\cup )\\ =A\cup B \\ by(identity\,law\,for\,\cap )\\$
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