Chapter 6.2, Problem 23ES

To determine

Fill in the blanks in the following proof that for all sets A and B, $\left(A-B\right)\cap \left(B-A\right)=\cancel{0}.$

Proof: Let A and B be any sets and suppose $\left(A-B\right)\cap \left(B-A\right)\cancel{=}\text{}\text{}\text{}\cancel{0}$. That is, suppose there were an element x in (a). By definition of $\left(b\right)$, $x\in A-Bandx\in \left(c\right)$. Then by definition of set difference, $x\in A$ and $x\notin B$ and $x\in \left(d\right)$ and $x\notin \left(e\right)$. In particular $x\in A$ and $x\notin \left(f\right)$, which is a contradiction. Hence [the supposition that $\left(A-B\right)\cap \left(B-A\right)\cancel{=}\cancel{0}$ is false, and so] (g).