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: 29

Answer

$For\,\,all\,\,sets\,\,A,\,B,\,and\,C,\\ (A \cup B) - C = (A - C) \cup (B - C).\\ Proof:\,Suppose\,A,\,B,\,and\,C\,are\,any\,sets.\\ Then\, (A \cup B) - C=(A\cup B)\cap C^c \\ by(set\,\,dif\! ference\,\,law)\\ =C^c\cap (A\cup B)\\ by(commutative\,\,law\,\,for\,\,\cap)\\ =(C^c\cap A)\cup (C^c\cap B)\\ by(distributive\,\,law)\\ =(A\cap C^c)\cup (B\cap C^c)\\ by(commutative\,\,law\,\,for\,\,\cap)\\ =(A-C)\cup (B-C)\\ by(set\,\,dif\! ference\,\,law)\\ $

Work Step by Step

$For\,\,all\,\,sets\,\,A,\,B,\,and\,C,\\ (A \cup B) - C = (A - C) \cup (B - C).\\ Proof:\,Suppose\,A,\,B,\,and\,C\,are\,any\,sets.\\ Then\, (A \cup B) - C=(A\cup B)\cap C^c \\ by(set\,\,dif\! ference\,\,law)\\ =C^c\cap (A\cup B)\\ by(commutative\,\,law\,\,for\,\,\cap)\\ =(C^c\cap A)\cup (C^c\cap B)\\ by(distributive\,\,law)\\ =(A\cap C^c)\cup (B\cap C^c)\\ by(commutative\,\,law\,\,for\,\,\cap)\\ =(A-C)\cup (B-C)\\ by(set\,\,dif\! ference\,\,law)\\ $
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