The difference of two set differences $(A-C) - (B-C)$ I am trying to write the statement above using logic. I managed to break up the first difference sign using the definition of set difference and get: 
Suppose $$x\in (A-C) \land x\notin(B-C),$$ 
but I have no idea how to break it further.
 A: @amWhy has an excellent answer to this, but as an alternative you can also work directly with the set operators:
$$(A - C) - (B - C) = $$
$$(A \cap C^C) \cap (B \cap C^C)^C=$$
$$A \cap C^C \cap (B^C \cup C)=$$
$$A \cap C^C \cap B^C=$$
$$A \cap (C \cup B)^C=$$
$$A - (C \cup B)$$
A: $x \in (A-C)-(B-C)\tag{(1) given}$
$\iff x \in A \land x \notin C \land \lnot(x \in B \land x \notin C)\tag{(2) def. set-difference}$
$\iff x\in A \land x \notin C \land (x \notin B \lor x \in C)\tag{(3) DeMorgan's}$
$\iff (x \in A \land x \notin C \land x \notin B) \lor\; \underbrace{\color{grey} { ( x \in A \land x \notin C \land x \in C)}}_{\large\varnothing} \tag{(4) distribution}$  
$$ \iff x \in A \land x \notin C \land x \notin B\tag{(5)$\color{grey}{\text{Contradiction in} (4)}$}$$
$$ \iff x\in A \land x \notin B \land x\notin C\tag{(5) commutativity}$$
$\iff x\in A \land \lnot (x \in B \lor x \in C)\tag{(6) DeMorgan's }$
$\iff x \in A \land x \notin (B\cup C)\tag{(7) def. Set union}$
$$\iff x \in A - (B\cup C)\tag{(8), def. set difference}$$
Hence $$(A-C)- (B-C) = A-(B\cup C)$$
