Chapter 6.2, Problem 27ES

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

Proof: Let A and B be any sets and suppose $\left(A-B\right)\cap \left(B-A\right)\ne \varnothing .$ That is, suppose there is an element x in (a) . By definition of (b) $x\epsilon A-B$ and $x\epsilon A-B$ and $x\epsilon$ . (c) Then by definition of set difference, $x\epsilon A$ 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 $(A-B)\cap \left(B-A\right)\ne \varnothing$ is false, and so] (g)