Is $A\cap B$ the same as $A\cup B-(A-B)-(B-A)$? Is $A\cap B$ the same as $A\cup B-(A-B)-(B-A)$?
I know that $A\cap B$ is equivalent to $A-(A-B)$ but I got to the first answer on my own and I want to know if it is correct.
 A: Yes, it is the same, indeed it results that
$A\cap B=(A\cup B)-(A-B)-(B-A)\;.\quad\color{blue}{(*)}$
Now we are going to prove the equality $(*)$.
For all $\;x\in A\cap B\;,\;$ it follows that $\;x\in A\;$ and $\;x\in B\;,$
hence $\;x\in A\cup B\;,\;$ but $\;x\notin A-B\;$ and $\;x\notin B-A\;,$
indeed if $\;x\in A-B\;$ or $\;x\in B-A\;,\;$ then $\;x\notin B\;$ or $\;x\notin A\;,\;$ but it is impossible because $\;x\in A\cap B\;.$
Therefore $\;x\in A\cup B\;,\;x\notin A-B\;,\;x\notin B-A\;,\;$ consequently we get that
$x\in (A\cup B)-(A-B)-(B-A)\;.$
Conversely, for all $\;x\in (A\cup B)-(A-B)-(B-A)\;,$
it follows that $\;x\in A\cup B\;,\;x\notin A-B\;$ and $\;x\notin B-A\;,$
hence $\;x\in A\;$ or $\;x\in B\;.$
If $\;x\in A\;,\;$ then $\;x\in B\;$ because $\;x\notin A-B\;,\;$ moreover
if $\;x\in B\;,\;$ then $\;x\in A\;$ because $\;x\notin B-A\;.$
So in any case it results that $\;x\in A\;$ and $\;x\in B\;,\;$ hence
$x\in A\cap B\;.$
A: Set difference is not associative, so it would be better to add parentheses:
$$
A\cap B=\bigl((A\cup B)-(A-B)\bigr)-(B-A)
$$
Let's tackle the first parenthesis: if $U$ is some set that contains both $A$ and $B$, set $X^c=U-X$, for $X\subseteq U$. Then $A-B=A\cap B^c$ and therefore
\begin{align}
(A\cup B)-(A-B)
&=(A\cup B)\cap(A\cap B^c)^c \\[4px]
&=(A\cup B)\cap(A^c\cup B) && \text{(De Morgan)} \\[4px]
&=(A\cap A^c)\cup B && \text{(distributivity)} \\[4px]
&=B
\end{align}
Now we're almost done:
\begin{align}
\bigl((A\cup B)-(A-B)\bigr)-(B-A)
&=B\cap(B\cap A^c)^c \\[4px]
&=B\cap(B^c\cup A) && \text{(De Morgan)} \\[4px]
&=(B\cap B^c)\cup (B\cap A) && \text{(distributivity)} \\[4px]
&=B\cap A \\[4px]
&=A\cap B
\end{align}
A: Well, if $A\cap B$ is equivalent to $A-(A-B)$ then
switching the roles of $A$ and $B$:
$B\cap A$ is equivalent to $B-(B-A).$
$A \cap B = B \cap A$ and $A \cup A=A$ so
$$(A \cap B) \cup (B \cap A) = A \cap B$$
or
$$[A-(A-B) ]\cup [B-(B-A)]=(A\cup B)-(A-B)-(B-A) = A \cap B.$$
