Proof of $(A - B) - C = A - (B \cup C)$ I have several of these types of problems, and it would be great if I can get some help on one so I have a guide on how I can solve these.  I tried asking another problem, but it turned out to be a special case problem, so hopefully this one works out normally. 
The question is:
Prove $(A - B) - C = A - (B \cup C)$
I know I must prove both sides are not equivalent to each other to complete this proof. Here's my shot: We start with left side.


*

*if $x \in C$, then $x \notin A$ and $x \in B$.

*So $x \in (B \cup C)$

*So $A - (B \cup C)$


Is this the right idea? Should I then reverse the proof to prove it the other way around, or is that unnecessary? Should it be more formal?
Thanks!

In a variation of the currently approved answer, I would prove this through the following simple calculation: for every $x$,
$$
\begin{array}{ll}
& x \in (A - (B \cup C)) \\
\equiv & \;\;\;\text{"definition of $-$"} \\
& x \in A \land \lnot (x \in (B \cup C)) \\
\equiv & \;\;\;\text{"definition of $\cup$"} \\
& x \in A \land \lnot (x \in B \lor x \in C) \\
\equiv & \;\;\;\text{"logic"} \\
& x \in A \land \lnot (x \in B) \land \lnot (x \in C) \\
\equiv & \;\;\;\text{"definition of $-$, to work towards our goal"} \\
& x \in (A - B) \land \lnot (x \in C) \\
\equiv & \;\;\;\text{"definition of $-$"} \\
& x \in ((A - B) - C) \\
\end{array}
$$
By the definition of set equality, this proves the given theorem.
