Prove or disprove. Calculations on sets. Problem with A u -A Prove or disprove 
 $(A \triangle B) \setminus (A \setminus C) = (A \cap −B \cap C) ∪ (B \setminus A)$
Using circles I have shown that they are equal, but now I have to prove it in more official way. So I am trying to get right side from the left one
$L = [(A − B) \cup (B − A)] \setminus (A \setminus C) = [(A \cap −B) \cup (B \cap −A)] \ (A \cap −C) 
= [(A \cup −B) \cup (B \cap − A)] \cap (−A \cup C) = [A \cup (B \cap −A) \cup −B \cup (B \cap −A)]  \cap (−A \cup C)$
And what now? I can write it in that form:
$(A \cup B) \cap (A \cup −A)$ ... , however what is $A \cup −A$? Is it $A \cup B \cup C$?
 A: Subtraction of a set is equivalent to an intersection with its complement:
$$X\setminus Y \equiv X\cap\overline Y$$
so the symmetric difference is
$$A \oplus B \equiv (A\setminus B)\cup(B\setminus A)\equiv (A\cap\overline B)\cup(\overline A\cap B)$$
Then
$$\begin{align}
(A \oplus B) \setminus  (A \setminus  C) &= ((A\cap\overline B)\cup(\overline A\cap B)) \cap \overline{A\setminus C} \\
 &= ((A\cap\overline B)\cup(\overline A\cap B)) \cap \overline{A\cap\overline C} \\
 &= ((A\cap\overline B)\cup(\overline A\cap B)) \cap (\overline A\cup C) \\
 &= (A\cap\overline B) \cap (\overline A\cup C)\cup(\overline A\cap B) \cap (\overline A\cup C) \\
 &= (A\cap\overline B \cap \overline A)\cup(A\cap\overline B \cap C)\cup(\overline A\cap B \cap \overline A)\cup(\overline A\cap B \cap C) \\
 &= \emptyset\cup(A\cap\overline B \cap C)\cup(\overline A\cap B \cap \overline A)\cup(\overline A\cap B \cap C) \\
 &= (A\cap\overline B \cap C)\cup(\overline A\cap B \cap \overline A)\cup(\overline A\cap B \cap \overline A)\cup(\overline A\cap B \cap C) \\
 &= (A\cap\overline B \cap C)\cup(B \cap \overline A)\cup(\overline A\cap B)\cup(\overline A\cap B \cap C) \\
 &= (A\cap\overline B \cap C)\cup(B \cap \overline A)\cup(\overline A\cap B\cap\Omega)\cup(\overline A\cap B \cap C) \\
 &= (A\cap\overline B \cap C)\cup(B \cap \overline A)\cup(\overline A\cap B\cap(\overline C\cup C))\cup(\overline A\cap B \cap C) \\
 &= (A\cap\overline B \cap C)\cup(B \cap \overline A)\cup(\overline A\cap B\cap\overline C)\cup(\overline A\cap B\cap C)\cup(\overline A\cap B \cap C) \\
 &= (A\cap\overline B \cap C)\cup(B \cap \overline A)\cup(\overline A\cap B\cap\overline C)\cup(\overline A\cap B\cap C) \\
 &= (A\cap\overline B \cap C)\cup(B \cap \overline A)\cup(\overline A\cap B\cap(\overline C\cup C)) \\
 &= (A\cap\overline B \cap C)\cup(B \cap \overline A)\cup(\overline A\cap B\cap\Omega) \\
 &= (A\cap\overline B \cap C)\cup(B \cap \overline A)\cup(\overline A\cap B) \\
 &= (A\cap\overline B \cap C)\cup(B \cap \overline A) \\
 &= (A \cap \overline B \cap C) \cup (B \setminus  A)
\end{align}$$
Q.E.D.
A: Actually, when writing $-B$ one should specify in advance what set the complement refers to. 
In this case it is implicit that the “universe” $U$ is a set containing $A$, $B$ and $C$; for the purpose of the problem, you can take as universe the set $U=A\cup B\cup C$.
It's probably better to first simplify the right-hand side. Set $X=A\cap-B$ and $Y=B\setminus A=B\cap-A$; then we have
$$
(A\cap -B\cap C)\cup(B\setminus A)=(X\cap C)\cup Y
=(X\cup Y)\cap(C\cup Y)
$$
by distributivity. 
By definition, $X\cup Y=A\mathbin{\triangle}B$; therefore the left-hand side can be rewritten as
$$
(A \mathbin{\triangle} B) \setminus (A \setminus C)=(X\cup Y)\cap-(A\cap -C)
$$
By De Morgan, $-(A\cap -C)=-A\cup C$, so we have
\begin{align}
(X\cup Y)\cap-(A\cap -C)
&=(X\cup Y)\cap(-A\cup C)\\
&=\bigl((X\cup Y)\cap-A\bigr)\cup\bigl((X\cup Y)\cap C\bigr)\\
&=\bigl((X\cap-A)\cup(Y\cap-A)\bigr)\cup\bigl((X\cup Y)\cap C\bigr)\\
&=\bigl(\emptyset\cup Y\bigr)\cup\bigl((X\cup Y)\cap C\bigr)\\
&=Y\cup\bigl((X\cup Y)\cap C\bigr)
\end{align}
because
$$
X\cap-A=A\cap-B\cap-A=\emptyset
$$
and
$$
Y\cap-A=B\cap-A\cap-A=B\cap-A=Y
$$
so we can go on and, finally,
\begin{align}
(A \mathbin{\triangle} B) \setminus (A \setminus C)
&=Y\cup\bigl((X\cup Y)\cap C\bigr)\\
&=\bigl(Y\cup(X\cup Y)\bigr)\cap(Y\cup C)\\
&=(Y\cup X)\cap(Y\cup C)\\
&=(X\cup Y)\cap(C\cup Y)
\end{align}
