How am I to interpret this result? Prove of give counter example: Let A and B be sets.

$$ A \backslash ( A \backslash B) = B \backslash ( B \backslash A) $$

An attempt at a proof:

If $ x \in  A \backslash ( A \backslash B)$ then $x \in A$ and $ x \notin A \backslash B$.

But if $ x \notin A \backslash B$ then either $x \notin A,B$ or $x \in B$ which is a contradiction. 

Am I wrong?  Is there an obvious counter example?
$$A\backslash (A\backslash B)=A\cap (A\cap B^c)^c=A\cap(A^c\cup B)=(A\cap A^c)\cup (A\cap B)=A\cap B,$$

$$B\backslash (B\backslash A)=B\cap (B\cap A^c)^c=B\cap(B^c\cup A)=(B\cap B^c)\cup (B\cap A)=B\cap A= A\cap B,$$

and thus both are equal.

To answer to your question:

There is no contradiction. If $x\in A\backslash (A\backslash B)$, then $x\in A$ but $x\notin A\backslash B$. Therefore $x\in B$ too. Indeed, if $x\notin A\backslash B$, then $x\in A^c$ or $x\in B$. But if $x\in A$, then $x$ can't be in $A^c$, therefore $x$ must be in $B$. We conclude that $x\in A$ and $x\in B$, what we can traduce by $x\in A\cap B$.
There is no contradiction. &nbsp; In point of order, $x\notin A\setminus B$ means either $x\notin A$ or $x\in B$. &nbsp; That is not an "and".

So if we take any $x$ in $A\setminus(A\setminus B)$, then we have "any $x$ in A except those not also in $B$".  Which is simply "any $x$ only in both $A$ and $B$".

$$A\setminus(A\setminus B) = A\cap B$$