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: 36

Answer

$For\,\,all\,\,sets\,\,A,\,B,\,and\,C,\\ ((A^c \cup B^c) - A)^c = A \\ proof:\\ ((A^c \cup B^c) - A)^c=((A^c \cup B^c) \cap A^c)^c \\ (set\,\,dif\! ference\,\,law )\\ =(A^c \cup B^c)^c \cup (A^c)^c \\ (De\,morgan\,\,law )\\ =((A^c)^c \cap (B^c)^c)\cup (A^c)^c \\ (De\,morgan\,\,law )\\ =(A\cap B)\cup A \\ (double\,\,complement\,\,law) \\ = A \cup (A\cap B) \\ (commutative\,\,law)\\ =A \\(absorption\,\,law)\\$

Work Step by Step

$For\,\,all\,\,sets\,\,A,\,B,\,and\,C,\\ ((A^c \cup B^c) - A)^c = A \\ proof:\\ ((A^c \cup B^c) - A)^c=((A^c \cup B^c) \cap A^c)^c \\ (set\,\,dif\! ference\,\,law )\\ =(A^c \cup B^c)^c \cup (A^c)^c \\ (De\,morgan\,\,law )\\ =((A^c)^c \cap (B^c)^c)\cup (A^c)^c \\ (De\,morgan\,\,law )\\ =(A\cap B)\cup A \\ (double\,\,complement\,\,law) \\ = A \cup (A\cap B) \\ (commutative\,\,law)\\ =A \\(absorption\,\,law)\\$
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