Prove that $A\setminus (B\setminus C) = (A\setminus B) \cup (A\setminus C^c)$ for sets $A,\ B,\ C$ in some Universal Set $U$. I'm working on this proof for some students I am tutoring and I've gotten a little stuck.   I want to show them how to do a proof in complete, extravagant detail and get them familiar with ''element chasing'' in set proofs for an intro to discrete class. I just got stuck and maybe I'm just too tired to see it.  Here's where I am so far.
Notation: $A^c$ is the complement of $A$.  $A\backslash B$ means $A \cap B^c$.

Prove that $A\setminus (B\setminus C) = (A\setminus B) \cup (A\setminus C^c)$ for sets $A,\ B,\ C$ in some Universal Set $U$.

Proof:
($\Leftarrow$)
Let $x \in (A \backslash B) \cup (A\backslash C^c)$.
This means that $x\in (A \cap B^c) \cup (A \cap C^{c^c})$.
Simplifying, this means that $x \in (A \cap B^c) \cup (A \cap C)$.
So $x\in (A \cap B^c)$ or $x\in (A\cap C)$.
Now we have two cases. 
Case 1: $  x \in A \cap B^c.$
So $x \in A$ and $ x\in B^c$.
Note: Now we must think outside the box a little bit.  We want to show that $x$ must be in $C$ but we don't have anything to do with $C$ in what we are able to derive from our assumptions.  We only have that $x$ is in $A$ and not in $B$.   This is where we must consider our problem.  We have the Universe $U$.  We know that $A,B,C$ are all in the universe.  So any point we chose in those sets must also be in $U$.   So think about this for a moment.  We don't know if our $x$ is in $C$ or not.  We need it to be in order to have the solution we are after.    So where is $x$ in relation  to $C$?   It's either in it or not in it.   So $x \in C$ or $x\in C^c$.  So we can examine each of these cases and we will find that if $x \in C^c$, we will get a contradiction.
Case 1.a.  Suppose $x \in C$.
Then $x \in A$ and $x \in B^c$ and $x \in C$.
So $x \in A$ and $ x\in  (B^c \cap C)$.
So $x \in A \cap (B^c \cap C)$.
So $x \in A \cap (B \cap C^c)^c$.
So $x \in A \cap (B\backslash C)^c$.
So $x \in A \backslash (B \backslash C)$.
Case 1.b.  Suppose $x \in C^c$.
Then $x \in A$ and $x\in B^c$ and $x\in C^c$.
So $x \in A \cap C^c$ and $x \in A\cap B^c$.
Now I'm stuck.   I know that I need to develop a contradiction because $x$ cannot be in $C^c$, but I'm just not seeing it.  Any suggestions?   If I can see this one, I'll be able to see the similar method I need to develop for case 2: $x\in A \cap C$ where I need to examine whether $x \in B$ or $B^c$.
 A: In case 1: $$x\in A\cap B^c\\x\in A,x\in B^c\\x\in A,x\not\in B\\x\in A,x\not\in B\cap C^c\\x\in A,x\not\in B\backslash C\\x\in A,x\in(B\backslash C)^c\\x\in A\cap(B\backslash C)^c\\x\in A\backslash(B\backslash C)$$
You don't need subcases (in case 1) for $x\in C, x\not\in C$.
(Incidentally, case 2 is as simple: there, you don't need subcases for $x\in B,x\not\in B$.)
A: There is yet another solution, namely you can rewrite 
$$A\setminus (B\setminus C) = (A\setminus B) \cup (A\setminus C^c)$$
using
$$X - Y = X \cap Y^c$$
into
$$A\cap \color{blue}{(B\cap C^c)^c} = (A\cap \color{red}{B^c}) \color{red}{\cup} (A\cap \color{red}{C}).$$
Now it is enough to use De Morgan law for blue part of LHS and the distributive law for red part of RHS:
$$A\cap \color{blue}{(B^c\cup C)} = A \cap (\color{red}{B^c \cup C}).$$
I hope this helps $\ddot\smile$
A: In your proof, there exists an error. Note the following fact : 
$x \in A \cap (B^c \cup C)$ iff
$x \in A \cap (B \cap C^c)^c$.
A: it might be enlightening for your students to do this 'algebraically':
If you go to the indicator functions and the set $\mathbf{2}^U=\mathbb{F}_2^U$ then
The symmetric difference of $a$ and $b$ is a+b
The complement of $a$ is $1+a$
The intersection of $a$ and $b$ is $ab$ 
The union of $a$ and $b$ is $U(a,b)=a+b+ab$
$a$ set-minus $b$ is $a(1+b)$
So the LHS
$$
a(1+(b(1+c))=a+ab(1+c)=a+ab+abc
$$
and the RHS
$$
U(a(1+b),a(1+1+c))=U(a+ab,ac)=a+ab+ac+abc+aac=a+ab+abc
$$
(note that $x^2=x$ and $x+x=0$)
A: Here is how I would present this proof.
The approach I learned is to start with the most complex side, calculate the elements of that set using the definitions, and then simplify, and finally work towards the other side of the equality.
In other words, for all $\;x\;$,
\begin{align}
& x \in (A \setminus B) \cup (A \setminus C^c) \\
\equiv & \qquad \text{"definition of $\;\cup\;$"} \\
& x \in A \setminus B \;\lor\; x \in A \setminus C^c \\
\equiv & \qquad \text{"definition of $\;\setminus\;$, twice; definition of $\;^c\;$"} \\
& (x \in A \land \lnot(x \in B)) \;\lor\; (x \in A \land \lnot(\lnot(x \in C))) \\
\equiv & \qquad \text{"logic, simplify: double negation, factor out common conjunct"} \\
& x \in A \land (\lnot(x \in B) \lor x \in C) \\
\equiv & \qquad \text{"logic: DeMorgan -- to work towards $\;x \in A \land \lnot \ldots\;$, suggested by goal"} \\
& x \in A \land \lnot (x \in B \land \lnot(x \in C)) \\
\equiv & \qquad \text{"definition of $\;\setminus\;$, twice"} \\
& x \in A \setminus (B \setminus C) \\
\end{align}
This proves the equality by set extensionality.
The only things needed for this type of proof are knowledge of the definitions, and of the laws of (propositional/predicate) logic.
