Proposed proof of set theoretic result I am tasked with proving the following:
$$ (A - B)\cap (B-A) = \varnothing $$
My Attempt:
Suppose there exist a $x \in (A - B)\cap (B-A) $ then:
 \begin{align*}
    x \in (A - B)\cap (B-A) &\iff (x \in A \land x \notin B) \land  (x \in B \land x \notin  A) \\
    &\iff  x \in A \land ( x\notin B \land  x  \in B) \land x \notin A \\
    &\iff  x \in A \land ( x\in B \land  x  \notin B) \land x \notin A \\
    &\iff  (x  \in A  \land x \notin A) \land ( x\in B \land  x  \notin B)\\
    &\iff  x \in (A-A) \cap (B-B)
  \end{align*}
Since $B-B = \varnothing $ and $A-A = \varnothing$ 
then $x \in  \varnothing $. An contradiction 
Is this attempt  correct? Can it be improved in anyway?
 A: Your proof is fine.
A shorter proof could go by noting that if there is an $x\in (A-B)\cap (B-A)$ then $x\in A-B$ and $x\in B-A$. The first will imply $x\in A$ while the second $x\not\in A$, so we'll have a contradiction.
A: You could simplify your proof by using exactly the same idea but writing it out in terms of set algebra instead of using "$x\in\cdots\,$" all the time.  Thus:
$$(A - B)\cap (B-A)=A\cap \overline B\cap B\cap\overline A=(A\cap\overline A)\cap(B\cap\overline B)=\varnothing\cap\varnothing=\varnothing\ .$$
A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
$Here is a simpler version of your proof: for all $\;x\;$,
$$\calc
x \in (A - B) \cap (B - A)
\calcop{\iff}{definition of $\;\cap\;$}
x \in A - B \;\land\; x \in B - A
\calcop{\iff}{definition of $\;-\;$, twice; parentheses can be left out}
x \in A \land x \not\in B \;\land\; x \in B \land x \not\in A
\calcop{\iff}{logic: contradiction}
x \in A \land \text{false} \land x \not\in A
\calcop{\iff}{logic: simplify}
\text{false}
\endcalc$$
In other words, by the definition of $\;\varnothing\;$, $\;(A - B) \cap (B - A) = \varnothing\;$.
