Fill in the blanks in the following proof that for all sets A and B, (A B) (B A) =

Chapter 6, Problem 24

(choose chapter or problem)

Fill in the blanks in the following proof that for all sets A and B, \((A-B) \cap(B-A)=\emptyset\).

Proof: Let A and B be any sets and supppose \((A-B) \cap (B-A) \neq \emptyset\). That is, suppose there were an element x in   (a)  . By definition of   (b)  , \(x \in A-B\) and \(x \in\)   (c)  . Then by definition of set difference, \(x \in A\) and \(x \notin B\) and \(x \in\)   (d)   and \(x \notin\) (e)  . In particular \(x \in A\) and \(x \notin\)   (f)  , which is a contradiction. Hence [the supposition that \((A-B) \cap(B-A) \neq \emptyset\) is false, and so]   (g)  .

Text transcription:

(A-B) cap(B-A)= emptyset

(A-B) cap (B-A) neq emptyset

x in A-B

x in

x in A

x notin B

x in

x notin

x in A

x notin

(A-B) cap(B-A) neq emptyset