Fill the blank to make a true identity I need to make a true identity by filling the missing part here:
 $$C\setminus(A\triangle B) = (A\cap C)\triangle ...................$$
I used Venn diagrams to find a solution which led me to $(A\cap C)\triangle(C\setminus B)$, however when I tried to represent what it means for x to be an element of both sides of the solution. Both answer differs, or I missing something. Here is what I have done:
For $x$ to be an element of $C\setminus(A\triangle B)$, means:
$$(x\in C\setminus x\in((A\setminus B)\cup(B\setminus A))) - \text{definition of }\triangle
\\
(x\in C\land\neg((x\in A\land x\notin B)\cup(x\in B\land x\notin A))) -\text{definition of }\setminus\\(x\in C\land\neg(x\in A\land x\notin B)\lor(x\in B\land x\notin A)) - \text{definition of }\cup\\(x\in C\land(\neg(x\in A\land x\notin B)\land \neg(x\in B\land x\notin A))) - \text{DeMorgan's law}\\(x\in C\land((x\notin A\lor x\in B)\land(x\notin B\lor x\in A)))-\text{DeMorgan's law}\\
\text{after associative law I end up  with }  x\in C\land(x\notin A\lor x\in B)\land(x\notin B\lor x\in A)$$
The right side $(A\cap C)\triangle(C\setminus B):$
$$((A\cap C)\setminus(C\setminus B))\cup((C\setminus B)\setminus(A\cap C)) \text{ means:}\\ ((x\in A\land x\in C)\setminus x\in(C\setminus B))\cup(x\in(C\setminus B)\setminus(x\in A\land x\in C))-\text{definition of }\cap\\((x\in A\land x\in C)\land\neg(x\in C\land x\notin B))\cup((x\in C\land x\notin B)\land\neg(x\in A\land x\in C))- \text{definition}\ \setminus\\((x\in A\land x\in C)\land\neg(x\in C\land x\notin B))\cup((x\in C\land x\notin B)\land\neg(x\in A\land x\in C))-\text{definition}\ \cup\\((x\in A\land x\in C)\land(x\notin C\lor x\in B))\lor((x\in C\land x\notin B)\land(x\notin A\lor x\notin C))-\text{DeMorgan's\ law}\\(x\in A\land (x\in C\land(x\notin C\lor x\in B))\lor((x\in C\land x\notin B)\land(x\notin A\lor x\notin C))-\text{associative law} \\(x\in A\land ((x\in C\land x\notin C)\lor (x\in C\land x\in B))\lor((x\in C\land x\notin B)\land(x\notin A\lor x\notin C))-\text{distributive law}\\(x\in A\land (x\in C\land x\in B))\lor((x\in C\land x\notin B)\land(x\notin A\lor x\notin C))-\text{contradiction law}\\(x\in A\land (x\in C\land x\in B))\lor( x\notin B\land(x\in C\land(x\notin A\lor x\notin C)))-\text{associative law}\\(x\in A\land (x\in C\land x\in B))\lor( x\notin B\land((x\in C\land x\notin A)\lor (x\in C\land x\notin C)))-\text{distributive law}\\(x\in A\land (x\in C\land x\in B))\lor( x\notin B\land(x\in C\land x\notin A))-\text{contradiction law}\\(x\in A\land x\in C\land x\in B)\lor( x\notin B\land x\in C\land x\notin A)-\text{associative law}\\x\in C\land ((x\in A\land x\in B)\lor( x\notin B\land x\notin A))$$
Sorry for being so verbose here. The question is are those results correct? If not - what I am missing? Thanks.
 A: I don't see any error in the calculations in the OP. It is true that $C \smallsetminus (A \triangle B) = (A \cap C) \triangle (C \smallsetminus B)$, in particular $A \cap B \cap C$ is included in both the right-hand and left-hand sides. 
As you correctly derived in the OP, $x \in C \smallsetminus (A \triangle B)$ is equivalent to 
\begin{align}\tag{1}
x\in C\land (x\notin A\lor x\in B)\land(x\notin B\lor x\in A),
\end{align}
whereas $x \in (A \cap C) \triangle (C \smallsetminus B)$ is equivalent to 
\begin{align}\tag{2}
x \in C\land ((x\in A\land x\in B)\lor( x\notin B\land x\notin A)).
\end{align}
We need to show that $(1)$ is equivalent to $(2)$. Clearly, it is enough to prove that $(x\notin A\lor x\in B)\land(x\notin B\lor x\in A)$ and $(x\in A\land x\in B)\lor( x\notin B\land x\notin A)$ are equivalent. 
\begin{align}
&(x\notin A\lor x\in B) \land (x\notin B\lor x\in A) \\
\iff& (x\notin B\lor x\in A) \land (x\notin A\lor x\in B) &\text{commutativity} \\
\iff& \big((x \in A \lor x \notin B) \land (x\in B \lor x \notin B)\big) \land \big((x \in A \lor x \notin A) \land (x\in B \lor x \notin A)\big)&\text{tautology added in }\land \\
\iff& \big((x \in A \land x\in B) \lor x \notin B\big) \land \big((x \in A \land x\in B) \lor x \notin A\big)&\text{distributivity} \\
\iff& (x\in A\land x\in B) \lor (x\notin B\land x\notin A) &\text{distributivity}
\end{align}
