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: 32

Answer

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

Work Step by Step

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