When is the set statement: (A\B)⊕(A ∩ B) = A true? "When is the set statement:

(A\B)⊕(A ∩ B) = A

true? Is it sometimes true, never true, or always true? If sometimes, state the specific cases where it is. A & B are arbitrarily selected sets."
I said it was sometimes, as, it seems if A is empty and B is empty, then A\B is empty, A ∩ B is empty, and so the entire left side is then empty, and so is A. Would that be correct? Further, are there are other cases where this is a true statement?
Thank you for your time. ^^
 A: If $A=\varnothing$, it doesn’t matter what $B$ is: $A\setminus B=\varnothing\setminus B=\varnothing$, and $A\cap B=\varnothing\cap B=\varnothing$, so $(A\setminus B)\oplus(A\cap B)=\varnothing\oplus\varnothing=\varnothing=A$.
But in fact it’s always true, for any sets $A$ and $B$. If $x\in A\cap B$, then $x\in B$, so $x\notin A\setminus B$. Conversely, if $x\in A\setminus B$, then $x\notin B$, so $x\notin A\cap B$. Thus, 
$$\begin{align*}
(A\setminus B)\oplus(A\cap B)&=\Big((A\setminus B)\setminus(A\cap B)\Big)\cup\Big((A\cap B)\setminus(A\setminus B)\Big)\\
&=(A\setminus B)\cup(A\cap B)\\
&=A\;.
\end{align*}$$
If you’re in doubt about that last step, notice that $A\setminus B$ consists of the things that are in $A$ but not in $B$, while $A\cap B$ consists of those things that are in both $A$ and $B$, so between them they pick up every element of $A$.
More generally, the symmetric difference of two disjoint sets is always their union:
if $X\cap Y=\varnothing$, then $X\oplus Y=X\cup Y$. Here the disjoint sets are $A\setminus B$ and $A\cap B$, and their union is $A$.
A: We can easily demonstrate this by starting with the most complex (left hand) side, expanding the definitions (see related answer), and simplifying: for every $\;x\;$,
\begin{align}
& x \in (A \setminus B) \oplus (A \cap B) \\
\equiv & \;\;\;\;\;\text{"definition of $\;\oplus\;$"} \\
& x \in A \setminus B \;\not\equiv\; x \in A \cap B \\
\equiv & \;\;\;\;\;\text{"definitions of $\;\setminus, \cap\;$"} \\
& x \in A \land x \not\in B \;\not\equiv\; x \in A \land x \in B \\
\equiv & \;\;\;\;\;\text{"move $\;\lnot\;$ to the outside -- to prepare for the next step"} \\
& \lnot(x \in A \land x \not\in B \;\equiv\; x \in A \land x \in B) \\
\equiv & \;\;\;\;\;\text{"factor $\;x \in A\;$ out of $\;\equiv\;$"} \\
& \lnot(x \in A \;\Rightarrow\; (x \not\in B \;\equiv\; x \in B)) \\
\equiv & \;\;\;\;\;\text{"logic: contradiction"} \\
& \lnot(x \in A \;\Rightarrow\; \text{false}) \\
\equiv & \;\;\;\;\;\text{"logic: simplify"} \\
& x \in A \\
\end{align}
By set extensionality, this proves the statement in question.
A: let $A,B \subset X$ and denote $(\forall Y), \bar Y=X \setminus Y$, then
$$(A\setminus B)\oplus(A\cap B) = (A \cap \bar B)\oplus(A\cap B)
= A \cap(\bar B \oplus B) = A \cap X = A
$$
