Prove that $B \setminus (B \setminus A) = A$ 
Prove that $B \setminus (B \setminus A) = A$

My conclusion is that  $B \setminus (B \setminus A) \neq A$, OR, to be precise, $B \setminus (B \setminus A) = A \cap B$.

If I'm wrong, please help me prove it. If I'm correct, how should I show that the question asked cannot be proved?
 A: You're correct. We can disprove via counterexample. Suppose that:
\begin{align*}
A &= \{1, 2\} \\
B &= \{2, 3\}
\end{align*}
Then observe that:
$$
B \setminus (B \setminus A)
= B \setminus \{3\}
= \{2\}
\neq \{1, 2\} = A
$$
A: Since $B\setminus X\subseteq B$, for every set $X$, a necessary condition for $B\setminus(B\setminus A)=A$ to hold is that $A\subseteq B$.
So just take $A$ any nonempty set and $B=\emptyset$ to get your counterexample.
Your conjecture is good: indeed $B\setminus(B\setminus A)=A\cap B$. You can prove it algebraically: if $U=A\cup B$ (or $U$ is any set containing both $A$ and $B$), we can denote $X'=U\setminus X$, when $X\subseteq U$. Then $B\setminus X=B\cap X'$ and so
\begin{align}
B\setminus(B\setminus A)
&= B\cap(B\setminus A)' && \text{(identity above)} \\
&= B\cap(B\cap A')' && \text{(identity above)} \\
&= B\cap(B'\cup A) && \text{(De Morgan)} \\
&= (B\cap B')\cup (B\cap A) && \text{(distributivity)} \\
&= \emptyset\cup(A\cap B) && \text{(commutativity)} \\
&= A\cap B
\end{align}
A: A truth table can be used as a logical Venn diagram. Each disjoint area in a Venn diagram corresponds to the appropriate member functions.
Using the definition $x\in B\setminus A\equiv x\in B \land x\notin A$ gives the following truth table.




$x\in A$
$x\in B$
$x\in B\setminus A$
$x\in B\setminus(B\setminus A)$
$x\in A\cap B$




$0$
$0$
$0$
$0$
$0$


$0$
$1$
$1$
$0$
$0$


$1$
$0$
$0$
$0$
$0$


$1$
$1$
$0$
$1$
$1$




As the last two columns are identical, we can conclude that $B\setminus(B\setminus A)\equiv A\cap B$.
A counter-example to the original conjecture is any $A\ne\emptyset$ such that $A\cap B=\emptyset$.
