Prove that $A \smallsetminus (A \smallsetminus B) = A \cap B$ $A$ and $B$ are any sets, prove that $A \smallsetminus (A \smallsetminus B) = A \cap B.$ This formula makes sense when represented on a Venn diagram, but I am having trouble with proving it mathematically.
I have tried letting $x$ be an element of $A$ and continue from there, but it doesn't seem to work out as a valid proof anyways. 
Could anyone please point me in the right direction?
Many thanks. 
 A: Let $x \in A \setminus (A \setminus B).$ 
Then  $x$ is an element such that $x \in A$ and $x \notin A \setminus B$. But if $x \notin A\setminus B$, with some additional work, we realize this implies that $x \in A$ and $ x \in B$. So $x \in A \cap B$.
Vice-versa: let $x \in A \cap B$ so $x \in A$ and $x \in B$. This implies  that $x \notin A \setminus B$. But given that $x \in A$ and $x \notin A \setminus B$ this implies that $x \in A \setminus (A \setminus B)$.
A: Notice that 
\begin{eqnarray*}
x\in A\setminus(A\setminus B) &\Leftrightarrow& (x\in A)\wedge \neg(x\in A\setminus B)\\
&\Leftrightarrow & (x\in A)\wedge  \neg((x\in A)\wedge \neg (x\in B))\\
&\Leftrightarrow & (x\in A)\wedge ((x\notin A)\lor(x\in B))\\
&\Leftrightarrow & ((x\in A)\wedge (x\notin A))\lor ((x\in A)\wedge(x\in B))\\
&\Leftrightarrow & (x\in A)\wedge (x\in B)\\
&\Leftrightarrow & x\in A\cap B.
\end{eqnarray*}
A: Did you know that 
$$A \smallsetminus B = A \cap \overline{B} \;\;?$$
$\begin{align} A \smallsetminus (A \smallsetminus B) &= A \smallsetminus (A \cap \overline{B}) \\ \\ &=  A \cap \overline{ (A \cap \overline{B})}\\ \\ &= A \cap (\overline{A} \cup B)\\ \\ &= (A \cap \overline{A} )\cup (A \cap B) \\ \\ &= A \cap B \end{align}$
A: Let $x\in A\cap B$.
Then $x\in A$ and $x\in B$.
Then $x\not \in A\setminus B$,
so $x\in A\setminus (A\setminus B)$.
Conversely,  if $x\in A\setminus (A\setminus B)$,
then $x\in A$ and $x\not \in (A\setminus B)$.
So $x\in A$ and $x\in B$.
That is $x\in A\cap B$...
