How do I prove set inequality: $(A-B)-C=(A-C)-(B-C)$ We are letting A,B, and C be non-empty sets. My issue with this is that I do not know how to prove it formally. Intuitively I just say let $x\in (A-B)-C$, so it is clear to me that $x$ is in $A$ but not in $B$ and then $x$ is in $A$ but not in $B$ and not in $C$. So I say $x\in(A-C)$ and $x\notin(B-C)$ which makes me jump to the conclusion $x\in(A-C)-(B-C)$ thus $(A-B)-C\subseteq(A-C)-(B-C)$.
The other way around is even more weird, I assume $x\in(A-C)-(B-C)$ AND intuitively I know that if I break it down I get that $x\in A,x\notin B,x\notin C$ but how do I get to this result I just end up writing the explanation. Is that perfectly fine? From there I say clearly $x\notin C$ and $x\in(A-B)$ so $x\in (A-B)-C$ and I have that they are both subsets of eachother hence equal.
 A: $(A-C)-(B-C)=(A\cap\bar C)\cap\overline{(B\cap\bar C)}=(A\cap\bar C)\cap(\bar B\cup C)=\underbrace{(A\cap\bar C\cap\bar B)}_{(A-B)-C}\cup\underbrace{(A\cap\underbrace{\bar C\cap C}_{\varnothing})}_{\varnothing}$
A: Turn $A-B$ into $A \cap\bar{B}$ and use De Morgan's rules.
A: Element chase:
If $x \in (A-B)-C$ then $x \in A$ and $x \not \in C$ so $x\in (A-C$.
And if $x \in (A-B) -C$ then $x \not \in B$ so $x \not \in B-C \subset B$.  So $x \in (A-C) -(B-C)$.
So $(A-B)-C \subset (A-C) - (B-C)$.
Likewise if $y \in (A-C)-(B-C)$ then $x \in A$.  $x \not \in B-C$ so if $x \in B$ then $x \in C$.  But $x \in A-C$ so $x \not \in C$.  So $x \not \in B$.  So $x \in A-B$.  And $x \not \in C$ so $x \in (A-B) -C$.
So $(A-C) -(B-C) \subset (A-B) -C$.
So $(A-B) - C = (A-C) -(B-C)$.
... Another way is to consider which elements are in which sets based on whether they are or are not in $A,B$ or $C$:
Take an element $x \in (A-B) -C$.
Is $x \in A$?  Yes.  Is $x\in B$? No.  Is $x \in C$? No.
Take an element $y \in (A-C) - (B-C)$.
Is $x \in A$?  Yes.  Is $x \in B$? Hmmm. Not if it's not in $C$ But it could be that it is in $B$ and $C$.  Is $x \in C$.  No.... Oh, wait, if it's not $C$ then it can't be in both $B$ and $C$ and since it can't be in $B-C$, it can't be in $B$ after all.
So in both cases $(A-B)-C$ and $(A-C)-(B-C)$ are both precisely the elements in $A$  that aren't in $B$ or $C$.
...
And if we want to do it formally:
$(A-B) - C = (A\cap B^c) \cap C^c = A\cap B^c \cap C^c$
whereas 
$(A-C) - (B-C) = (A\cap C^c)\cap (B\cap C^c)^c=$
$(A\cap C^c) \cap (B^c \cup C)= [(A\cap C^c) \cap B^c] \cup [(A\cap C^c) \cap C] = $
$(A\cap B^c \cap C^c) \cup [A\cap (C^c \cap C)]=$
$(A\cap B^c \cap C^c) \cup [A\cap \emptyset]=$
$(A\cap B^c \cap C^c) \cup \emptyset=$
$A\cap B^c \cap C^c$ also.
