TITLE: Using disjunction to prove that $A \setminus (A \setminus B) = A \cap B$
QUESTION [5 upvotes]: The problem is as follows:

Suppose $A$ and $B$ are sets. Prove that $A \backslash (A \backslash B) = A \cap B$.

I've rewritten the problem as a biconditional where $A \backslash (A \backslash B) \longleftrightarrow A \cap B$. Proving the first part, $A \backslash (A \backslash B)  \rightarrow A\cap B$, was easy enough. However, I'm finding the second part, $A\cap B \rightarrow A \backslash (A \backslash B)$, much harder. To prove that $x \in A \land (x \notin A \lor \ x \in B)$, all I have is that $x \in A$ and $x \in B$. Even more puzzling is that I'm suppose to be using a disjunction somewhere.    
I should be clear that I'm not simply trying to show an equivalence but write a logical proof.

REPLY [2 votes]: Proposition

$x \in A\backslash(A\backslash B) \leftrightarrow x \in A∩B$

Definitions needed
$A\backslash B = x \in A \wedge x \notin B$
$A\cap B = x \in A \wedge x \in B$

The Proof:
$$\begin{align} x \in A\backslash(A\backslash B) \leftrightarrow x \in A∩B & \equiv x \in A \wedge \neg (x \in A\backslash B) \leftrightarrow x \in A \wedge x \in B \\  
&\equiv x \in A \wedge \neg (x \in A \wedge \neg (x \in B)) \leftrightarrow x \in A \wedge x \in B     \\
&\equiv x \in A \wedge (x \notin A \vee x \in B) \leftrightarrow x \in A \wedge x \in B\\ \\
&\equiv (x \in A \wedge x \notin A) \vee (x \in A \wedge x \in B) \leftrightarrow x \in A \wedge x \in B \\ \\
&\equiv \bot \vee (x \in A \wedge x \in B) \leftrightarrow x \in A \wedge x \in B \\ \\
&\equiv x \in A \wedge x \in B \leftrightarrow x \in A \wedge x \in B \\ \\
\end{align}$$