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)