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!

(A-B)-C=A-(B ∪ C)
The right hand side means
A ∩ (B ∪ C)^c
(c means compliment)
Which, in turn, means, by deMorgan's law
A ∩ B^c ∩ C^c
Hence, by the associative property,
(A ∩ B^c) ∩ C^c
And, reinstating the minus sign,
(A-B)-C
Thus (A-B)-C=A-(B ∪ C)
