Solved: In 27 and 28 supply a reason for each step in the derivation
Chapter 6, Problem 28(choose chapter or problem)
For all sets A, B, and C, \((A \cup B)-(C-A)=A \cup(B-C)\).
Proof: Suppose A, B, and C are any sets. Then
\((A \cup B)-(C-A)=(A \cup B) \cap(C-A)^{c}\) by (a)
= \((A \cup B) \cap\left(C \cap A^{c}\right)^{c}\) by (b)
= \((A \cup B) \cap\left(A^{c} \cap C\right)^{c}\) by (c)
= \((A \cup B) \cap\left(\left(A^{c}\right)^{c} \cup C^{c}\right)\) by (d)
= \((A \cup B) \cap\left(A \cup C^{c}\right)\) by (e)
= \(A \cup\left(B \cap C^{C}\right)\) by (f)
= \(A \cup(B-C)\) by (g).
Text Transcription:
(A cup B)-(C-A)=A cup(B-C)
(A cup B)-(C-A)=(A cup B) cap(C-A)^c
(A cup B) cap(C cap A^c)^c
(A cup B) cap(A^c cap C)^c
(A cup B) cap((A^c)^c cup C^c)
(A cup B) cap(A cup C^c)
A cup(B cap C^c)
A cup(B-C)
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