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: 30

Answer

$For\,\,all\,\,sets\,\,A,\,B,\,and\,C,\\ (A \cap B) \cup C = (A \cup C) \cap (B \cup C).\\ proof:\\ (A \cap B) \cup C =C\cup (A \cap B) \\ by(commutative\,\,law\,\,for\,\,\cup)\\ =(C\cup A)\cap (C\cup B)\\ by(distributive\,\,law)\\ =(A\cup C)\cap (B\cup C)\\ by(commutative\,\,law\,\,for\,\,\cup)\\ $

Work Step by Step

$For\,\,all\,\,sets\,\,A,\,B,\,and\,C,\\ (A \cap B) \cup C = (A \cup C) \cap (B \cup C).\\ proof:\\ (A \cap B) \cup C =C\cup (A \cap B) \\ by(commutative\,\,law\,\,for\,\,\cup)\\ =(C\cup A)\cap (C\cup B)\\ by(distributive\,\,law)\\ =(A\cup C)\cap (B\cup C)\\ by(commutative\,\,law\,\,for\,\,\cup)\\ $
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