Proving equality of sets using proof by contradiction I was wondering how can I prove the following conjecture using proof by contradiction. 
If $A-B = B-A$, then it must be the case that $A=B$.
I tried to assume that $A \ne B$ and tried to show that $A-B = B-A$ still holds. However, I'm not sure how to deduce this part. I am new to set theory and proofs, so any kind of help would be appreciated.
 A: I tried to assume that $A≠B$ and tried to show that $A−B=B−A$ still holds. However, I'm not sure how to deduce this part.
Suppose there is an $a$ such that $a\in A$ and $a\notin B,$ yet $A-B = B-A$
$a\in A-B, a\notin B-A$
$A-B \ne B-A$ 
Contradiction!
All $a$ in $A$ are also in $B.$
And by an identical argument all $b$ in $B$ are also in $A.$

Alternatively, and a little longer for clarity.


*

*Assume $A\smallsetminus B=B\smallsetminus A$. 


*

*Take an arbitrary $a$


*

*Assume $a\in A$. 


*

*Assume $a\notin B$. 

*$a\in A \wedge a\notin B$ by conjunction .

*$a\in A\smallsetminus B$ by definition of set minus. 

*$a\in B\smallsetminus A$ by the first assumption (and substitution).

*$a\in B\wedge a\notin A$ by definition of set minus.

*$a\in B$ by simplification.

*Contradiction! ($a\notin B$ and $a\in B$)


*$a\in B$ via proof by contradiction.


*$a\in A\to a\in B$ via conditional proof


*$\forall x~(x\in A\to x\in B)$ by universal generalisation.

*$A\subseteq B$ by definition of subset

*$B\subseteq A$ similarly.

*$A=B$
