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: 35

Answer

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

Work Step by Step

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