Prove or disprove: $A-(B-C) = (A-B)-C$ This is what I have so far: 

First, we will attempt to show $A-(B-C) \subseteq (A-C) - C.$
    Let $x \in A -(B-C)$. 
    Then $x\in A$ and $x \notin (B-C)$. By DeMorgan's we have that 
$$x \notin (B-C) = x \notin B \vee x \in C.$$
    We have that $x \in A \wedge (x \notin B \vee x \in C)$. 
    Then we have that $(x \in A \wedge x \notin B) \vee (x \in A \wedge x \in C)$.

So by definition of $\cup$ and $\cap$, we have $x \in (A-B) \cup (x\in (A \cap C))$. I'm kind of stuck here.

I can see a contradiction here in that LHS says $x \in C$ but the RHS says $x \notin C$. 

Can anyone tell me what to do from here? I'm not quite sure how to formally state that the statement is false. I feel like I'm not being complete here.
Let $A=B=C = \{1\}$. Then
$$
(A - B) - C = \emptyset,
$$
but
$$
A-(B-C) = \{1\} - \emptyset = \{1\} \neq \emptyset.
$$
Why would you start trying to prove some random identity without first testing it in some easily computed cases to see if it's right or wrong??? Because I like to look for easy solutions before hard ones, I always try plugging in zero for one of the variables, or in this case because we're dealing with sets, the empty set.

If $A=\emptyset$ then $A-(B-C)=(A-B)-C$ turns into $\emptyset=\emptyset,$ check.

If $C=\emptyset$ it turns into $A-B=A-B,$ check.

If $B=\emptyset$ it turns into $A=A-C.$ Now that looks fishy, doesn't it? What if $C=A,$ then we get $A=\emptyset.$ That sure isn't an identity . . .

Counterexample: Let $B=\emptyset$ and $A=C=\mathbb R$ (or any nonempty set). Then 
$$A-(B-C)=\mathbb R-(\emptyset-\mathbb R)=\mathbb R-\emptyset=\mathbb R$$
while
$$(A-B)-C=(\mathbb R-\emptyset)-\mathbb R=\mathbb R-\mathbb R=\emptyset.$$