For what conditions on sets $A$ and $B$ the statement $A - B = B - A$ holds? The obvious response is $A = B$. But I came up with another response. I can't figure out what's my bias.
Lets say 


*

*$S = A - B$

*$T = B - A$
Proving $S=T$ means proving $S \subseteq T$ and $T \subseteq S$
Let's start with $S \subseteq T$
For any $x \in S$ :


*

*$x \in A$

*$x \notin B$


if $S \subseteq T$, $x \in T$ and : 


*

*$x \in B$

*$x \notin A$


There is no $x$ giving satisfaction, so $A$ and $B$ are empty.
I prove $T \subseteq S$ by symmetry
My solution is : $A = B = \emptyset$
What's wrong ?
 A: Let us look at the flaw in your logic. We are given that $S \subset T$. Then, if $x \in S$, then $x \in A - B$, so $x \in A$, and $x \notin B$. But also, $x \in T$, which means that $x \in B$ and $x \notin A$. Hence, $x \in B$ and $x \notin B$, which is a contradiction. Hence, no such $x$ exists, so $S$ must be empty. Reversing the argument, we conclude that $T$ must be empty. In other words, $A \subset B$ and $B \subset A$, so $B=A$.
A: I tried your reasoning using predicate logic.  I'm posting the answer to show an alternative way of reasoning.
So, $x \in A-B$ if $x \in A$ but $x\not\in B$.  Equivalently, $x\in B-A$ if $x \in B$ but $x \not\in A$.  Thus, $A-B = B-A$ if
\begin{equation}
\forall x : ( x \in A \wedge x\not\in B \Longleftrightarrow x\not\in A\wedge x\in B )
\end{equation}
Remember $\varphi \Longleftrightarrow \psi$ means $(\varphi \wedge \psi) \vee (\neg \varphi \wedge \neg \psi )$.  But in this case, the left term of this re-written equivalence is $x \in A \wedge x\not\in B \wedge x\not\in A\wedge x\in B$ and it is obviously impossible (equal to $\bot$), so the only possibility for $x$ is 
\begin{align}
  & \neg(x \in A \wedge x\not\in B) \wedge \neg(x\not\in A\wedge x\in B) \\
  =& (x \not\in A \vee x\in B) \wedge (x\in A\vee x\not\in B) \\
  =& (x \not\in A \wedge x\not\in B) \vee (x\in A\wedge x\in B) \vee (x \in A \wedge x\not\in A) \vee (x\not\in B\wedge x\in B) \\
  =& (x \not\in A \wedge x\not\in B) \vee (x\in A\wedge x\in B) \\
\end{align}
In other words, there are no $x$ such that $A$ contains $x$ but not $B$ or $B$ contains $x$ but not $A$.  That is, $A$ and $B$ have the same elements, i.e. $A=B$.
