Proving the statement $A\setminus(A\setminus B)=B\setminus(B\setminus A)$ I'm trying to prove/disprove the following statement:
$$A \setminus (A \setminus B) = B \setminus (B \setminus A)$$
From what I gather, it is true, since a simple venn diagram check suggests the two are the same. 
However, I am stuck on how to prove it.
Note: I am also given the following equalities to use in my proof:
$$A \setminus B = A \cap B^\complement$$
$$(A \setminus B)^\complement = A^\complement \cup B$$
So far, I have tried converting the first part of the original statement using the given equalities into the following: $$ A \cap (A^\complement \cup B) $$ but I have no idea if I'm on the right path. I can't seem to find out how to show that the two are subsets of one another.
Any help would be greatly appreciated!
 A: See $A\setminus(A\setminus B)$ construction:

Swap A and B symbols to get essentialy the same result for $B\setminus(B\setminus A)$ hence the equality.
Algebraically the sets' subtraction is equivalent to an intersection with a complement:
$$X \setminus Y = X \cap Y^\complement$$
hence:
$$\begin{align}A\color{green}\setminus(A\color{red}\setminus B) & = A\color{green}\setminus(A \color{red}\cap B^\color{red}\complement) \\
 & = A\color{green}\cap(A \color{red}\cap B^\color{red}\complement)^\color{green}\complement \\
 & = A\cap(A^\complement \cup B) \\
 & = (A\cap A^\complement) \cup (A\cap B) \\
 & = \emptyset \cup (A\cap B) \\
 & = A\cap B
\end{align}$$
A: Let $x, y \in A, B.$ Then, $x,y \not\in A\setminus B$. So, $x,y \in A\setminus(A\setminus B)$.
Above point is symmetric. So, $x,y \in B\setminus(B\setminus A)$.
Next, if $x\in A, x\not\in B$, then, $x\in (A\setminus B)$, giving $x\not\in A\setminus(A\setminus B)$. And similar result we will get for right hand side by symmetry.
Hence, 
$A\setminus(A\setminus B) = B\setminus(B\setminus A)\equiv A\cap B$
