Show that $(A-B)-C \subseteq A-C$. \begin{align}
c. & \quad  (A-B)-C \subseteq A-C\\
& \quad x \in [(A-B)-C]\text{ is equivalent to }x\in A\wedge x\not\in B\wedge x\not\in C,\text{ by definition of difference. }\\
&\quad x\in A-C\text{ is equivalent to }x\in A\wedge x\not\in C,\text{ by definition of difference.} 
\end{align}
Basically, I'm not sure how to proceed. I can see this two ways. One with $(A - B) - C$ having at least one element, and on in which it has no elements. Should I show how the left side is a subset of the right in both cases? Or is there something more simple I could do here? Also, is there a named law for going from $x$ is in $A$ and $x$ is in $B$ therefore $x$ is in $A$? Thanks!
 A: So every $x \in (A - B) - C$ is in $A-C$, yes? (You've done most of the "element chasing" in your preliminary work, save for the conclusion), i.e.,
$x \in (A - B) - C \implies x \in (A - B) \land x\notin C \implies x\in A \land x \notin B \land x \notin C.$  From the fact that $x \in A \land x \notin B \land x \notin C$ it is sufficient to know that $x \in A \land x \notin C$ from which it follows that $ x \in (A - C)$.
Therefore, by definition of set inclusion 
$$(A-B)-C \subseteq (A - C)$$
We've shown that any $x$ in $(A - B) - C$, if there is such an $x$, must be in $A - C$. That's it. We are not trying to prove set EQUIVALENCE. 
If there is no $x \in (A - B) - C$, it still holds that $(A - B) - C) \subseteq (A - C)$, as the empty set is the subset of every set.
The fact that $x \notin B$ (from the first set) doesn't matter, since logically, from the fact that $p \land \lnot q \land \lnot r,$ we can conclude $p \land \lnot r$ (and leave $\lnot q$ alone, since when each of $p$, $\lnot q$, $\lnot r$ are true, so is any one or any two of them true.
A: First, begin with $(A-B)-C$ and proceed by a chain of equalities:
$(A-B)-C=(A \cap B^c) \cap C^c = (A \cap C^c) \cap B^c = (A - C) \cap B^c$
So we can work with the expression $(A - C) \cap B^c$ rather than the original one. Now in general (meaning: for all sets $X$ and $Y$), it holds that $$X \cap Y \subseteq X.$$ Thus $$(A - C) \cap B^c \subseteq A-C.$$
But recall that $(A-B)-C = (A - C) \cap B^c$. Thus
$$(A - B) - C \subseteq A-C.$$
EDIT: The moral of the story is to use intersections, unions and complements. These operations have nice properties, while set-subtraction does not.
