Show that $( A \setminus B) \oplus C = ( A \oplus C) \oplus ( A \cap B)$ 

*Use any method you wish to verify the following identities: $( A \setminus B) \oplus C = ( A \oplus C) \oplus ( A \cap B)$
I chose to define $( A \oplus C) = \psi$, $( A \cap B) = \phi$, and $(A \setminus B) = \pi$, trying to break down the expression; expression now looks like $ \pi \oplus C = \psi \oplus \phi$. LHS--abitrary choice--
\begin{align*}
\pi \oplus C &:= (\pi \setminus C) \cup ( C \setminus \pi) \\ &\equiv (x \in \pi \wedge \neg (x \in C)) \vee ( x \in C \wedge \neg ( x \in \pi)) \\ &\equiv ( x \in A \wedge \neg (x \in B) \wedge \neg (x \in C)) \vee( x \in C \wedge \neg (x \in A \wedge \neg ( x \in B))) .
 \end{align*}
Now the RHS: \begin{align*}
\psi \oplus \phi :=&( \psi \setminus \phi) \cup ( \phi \setminus \psi) \\ &\equiv ( x \in \psi \wedge \neg( x \in \phi)) \vee ( x \in \phi \wedge \neg ( x \in \psi)) \\ 
&\equiv ((((x \in A \wedge \neg ( x \in C)) \vee ( x \in C \wedge \neg( x \in A))) \wedge \neg ( x \in A \wedge x \in B)) \vee ((( x \in A \wedge x \in B) \wedge \neg ( x \in A \wedge \neg ( x \in C)) \vee ( x \in C \wedge \neg( x \in A))
\end{align*}
I stopped here. I tried looking at ways to simplify the LHS and RHS, but I didn't see anything. Did I make a mistake somewhere? or am I not recognizing a way to simplify? If you decide to reply, could you reply with a similar method like the one I tried using? Thanks, for future refrence. 
 A: Assume $x \in (A \setminus B) \oplus C$.  If $x \in C$ and $x \in A$, then also $x \in B$.  If that's the case, then $x \notin A \oplus C$ but $x \in A \cap B$.
If $x \in C$ but $x \notin A$, then $x \in A \oplus C$ and $x \notin A \cap B$.
If $x \notin C$, then $x \in A \setminus B$, so $x \in A \oplus C$ and $x \notin A \cap B$.
This proves $(A \setminus B) \oplus C \subseteq (A \oplus C) \oplus (A \cap B)$.
Conversely, assume $x \in (A \oplus C) \oplus (A \cap B)$.  If $x \in A \cap B$, then $x \notin A \oplus C$.  But we know $x \in A$, so it follows that also $x \in C$.  Thus, $x \in C$ but $x \notin A \setminus B$.
If $x \in A \oplus C$, then $x \notin A \cap B$.  If $x \in A$, then $x \in A \setminus B$ and also $x \notin C$.  If instead $x \in C$, then $x \notin A$, so $x \notin A \setminus B$.
This proves $(A \oplus C) \oplus (A \cap B) \subseteq (A \setminus B) \oplus C$, so each side is contained in the other and they are equal.
A: The trick is that 


*

*$A\setminus B=A\cap(U\oplus B)$, where $U$ is the ‘“universe”; in your case $U=A\cup B\cup C$ suffices.

*Intersection is distributive over disjoint union, that is, $X\cap(Y\oplus Z)=(X\cap Y)\oplus(X\cap Z)$.

*$\oplus$ is associative and commutative. 
Now apply these facts:
$$
(A\setminus B)\oplus C=(A\cap(U\oplus B))\oplus C=(A\cap U)\oplus (A\cap B)\oplus C
$$
and you’re done with the help of 3.
A: Since the question allowed any method, it is easiest to use indicator fucntions.
The indicator function on the global set $X$ is $\mathbf{1}_{A}:X\rightarrow \{0,1\}$, taking value 1 on elements $x\in A$, and 0 otherwise. 
It is easy to verify the identities:$$\mathbf{1}_{A\cap B} = \mathbf{1}_{A}\mathbf{1}_{B}, \ \ \ \ \ 
\mathbf{1}_{A\cup B} = \mathbf{1}_{A}+\mathbf{1}_{B}-\mathbf{1}_{A}\mathbf{1}_{B},$$
$$\mathbf{1}_\overline{A} = \mathbf{1}_{X} - \mathbf{1}_{A}, \ \ \ \ \ 
\mathbf{1}_{A\oplus B} \equiv \mathbf{1}_{A}+\mathbf{1}_{B} \ \textrm{mod} \ 2.$$
Thus, we can calculate that the LHS is 
$$\mathbf{1}_{A}(\mathbf{1}_{X} - \mathbf{1}_{B}) + \mathbf{1}_{C} \ \textrm{mod 2},$$
and that the RHS is 
$$\big((\mathbf{1}_{A}+\mathbf{1}_{C})\ \textrm{mod 2} + \mathbf{1}_{A}\mathbf{1}_{B}\big) \ \textrm{mod 2}.$$
Since these are congruent modulo 2, and noting that valid indicator functions only take boolean (0, 1) values, these are the same indicator functions (i.e. they are identically equal on all elements in the global set). $\blacksquare$
