What can be equivalent to $x \in \overline{A - (C-B)}$ in this demonstration? 
Demonstrate
$\overline{A \triangle C -B} = B \cup (A \cap C) \cup (\overline{A} \cap \overline{C})$

To prove this, I first need to prove that $\overline{A \triangle C -B} \subseteq B \cup (A \cap C) \cup (\overline{A} \cap \overline{C})$.
$\subseteq$
Given an arbitrary element $x$:
$x \in \overline{A \triangle C -B}$
$x \in \overline{A-(C-B) \cup (C-B) - A}$
$x \in \overline{A-(C-B)} \cap \overline{(C-B) - A}$
$x \in \overline{A - (C-B)} \land x \in \overline{(C-B) - A}$
Now, I am a bit unsure about how to proceed. I have $x \in \overline{A - (C-B)}$, but, what can this be converted to? I wondered if it was equivalent to $x \in \overline{A} \land x \notin \overline{(C-B)}$ but I have the feeling that it isn't. How can I proceed?
 A: $C - B = C\cap \bar{B}$, and similarly $A - (C-B) = A \cap \overline{(C \cap \overline{B})}$.   Using De Morgan's laws:
$$
A \cap \overline{(C \cap \overline{B})} 
= A\cap (\overline{C} \cup B)
$$
Then complement over the whole thing is
$$
A\cup (C\cap\overline{B})
$$
again, using De Morgan's laws.  Then A can distribute into the expression to get an intersection of two unions, or you can rewrite it as
$$
A\cup (C - B)
$$
.  Whichever way is convenient for you problem.
A: To solve this on the element-level using logic, just peel away the definitions systematically, from the outside working your way in, and then simplify: for every $\;x\;$,
\begin{align}
& x \in \overline{A \triangle C - B} \\
\equiv & \;\;\;\;\;\text{"definition of $\;\overline{\phantom A}\;$"} \\
& \lnot(x \in A \triangle C - B) \\
\equiv & \;\;\;\;\;\text{"definition of $\;-\;$"} \\
& \lnot(x \in A \triangle C \land \lnot(x \in B)) \\
\equiv & \;\;\;\;\;\text{"definition of $\;\triangle\;$"} \\
& \lnot((x \in A \not\equiv x \in C) \land \lnot(x \in B)) \\
\equiv & \;\;\;\;\;\text{"logic: simplify using DeMorgan"} \\
& (x \in A \equiv x \in C) \lor x \in B \\
\end{align}
Now do the same for the right hand side, apply one of the definitions of logical equivalence ($\;\equiv\;$), and the proof is complete.
