Is correct this proof ?
$A\ \Delta \ B=(A\ ∪\ B)-(B\ ∩\ A)$
\begin{split}
x \in (A \ \cup B)- (B\ ∩\ A) & \Rightarrow x\in A \ \cup B\ \wedge x \notin A\ \cap \ B\ \\
& \Rightarrow (x \in A\ \vee x\in B) \wedge\ x\notin A\ \cap \ B\ \\ & \Rightarrow (x \in A \ \wedge x\notin A\cap B ) \ \vee \ (x \in B \ \wedge \ x\notin A\ \cap B\ )\\ &\Rightarrow (x \in A \ \wedge x\notin B ) \ \vee \ (x \in B \ \wedge \ x\notin A) \\&\Rightarrow\ x\in \boldsymbol{(A-B)\ \cup \ (B-A)}
\end{split}
Second part
$ x \in (A-B) \cup \ (B-A) \Rightarrow x\in (A – B)\ \vee x \in (B- A)$
\begin{split}
x\in (A – B) & \Rightarrow x \in A\ \wedge x\notin B\\ & \Rightarrow x \in A \cup B \ \wedge x\notin B \\ &\Rightarrow x \in A \cup B \ \wedge x\notin A
\cap B \ \\&\Rightarrow\ x\in \boldsymbol{(A\cup B)\ – \ (A\cap B)}
\end{split}
Best Answer
Yes, it's correct, but (was) incomplete; you need(ed) to prove the reverse inclusion. Just replace all your implications (in your first half) with biconditionals.