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!

Working with the Characteristic function of a set makes these problems easy:
$$1_{(A - B) - C}= 1_{A-B} - 1_{A-B}1_C=(1_A- 1_A1_B)-1_A1_C+ 1A1_B1_C \,$$
$$1_{A-(B \cup C)}= 1_{A}- 1_{A}1_{B \cup C}=1_A- 1_A ( 1_B+ 1_C -1_B1_C)\,$$
It is easy now to see that the RHS are equal.
