verfiy my proof (Prove that for all sets A, B, C, $(A \cup B)-C = (A-C) \cup (B-C)$) Prove that for all sets A, B, C, $(A \cup B)-C = (A-C) \cup (B-C)$
suppose $x \in A,B,C$
if $x\in A$ or $B$ or both , then by definition of union, $x\in A\cup B$.
Since $x\in (A \cup B) $ and $x\in C$, by definition of subtraction, $x\notin (A \cup B)-C$ 
Since $x \in A$ and $x \in C$, by definition of subtraction, $x \notin (A-C)$, the same is true if $x \in B$, then $x \notin (B-C)$.
Now if $x \notin (A-C)$ and $x \notin (B-C)$ then $x \notin (A-C) \cup (B-C)$ and $x \notin (A \cup B) -C$
thus $(A \cup B)-C = (A-C)\cup (B-C)$
verify my proof?
-thanks
 A: It's a bit confusing because you start by assuming "$x\in A,B,C$", but what does that mean? Are we assuming $x\in A\cap B\cap C$? If so, you should say that.
It's also not clear how it benefits us to assume that $x$ is in that intersection. There are precisely two things we need to show: (1) That every $x$ that's an element of the left-hand-side is also an element of the right-hand-side, and (2) That every $x$ that's an element of the right-hand-side is also an element of the left-hand-side. That's our way to show set equality.
Therefore, our first step should be to assume that $x$ is on one side or the other, and then prove that it's on the other side as well.
I'll show you one direction:

Suppose $x\in (A\cup B)-C$. Then $x\in A\cup B$ but $x\not\in C$. We have two cases (a) suppose $x\in A$. Then we have $x\in A-C$. (b) suppose $x\not\in A$, then $x\in B$. Thus, $x\in B-C$. In either case, we have $x\in(A-C)\cup(B-C)$.
We conclude: $(A\cup B)-C \subseteq (A-C)\cup(B-C)$

Do you see how I started with $x$ an element of the LHS, and ended up showing that the same $x$ must be an element of the RHS? Try and write the other direction, where you start with something in $(A-C)\cup(B-C)$, and show that it must also be in $(A\cup B)-C$.
A: I thought it might be useful to comment on the proof a little more. I don't think you even proved what you wanted to prove at all! 
I'll write the left-hand side set as $L = (A\cup B) - C$ and the right-hand side set as $R = (A - C)\cup (B - C)$. 
You showed that if $x\in A, B, C$, then $x\not\in L, R$. That's something, but that's not enough to conclude that $L = R$. 
What if we have $x\in R$ but $x\not\in L$? Your proof doesn't really eliminate that possibility. 
The safest approach to doing these proofs is to stick with the definitions. The way I define equality between sets is that I say two sets are equal whenever $x\in L$ if and only if $x\in R$. 
This means there are two parts: (1) I want to suppose $x\in L$ and then show that this implies $x\in R$, and then (2) I want to suppose $x\in R$ and show that this implies $x\in L$. Once you do that you have that $x\in L$ if and only if $x\in R$, and thus $L = R$ by definition.  
A: after rethinking my proof i came up with:
suppose $x \in (A \cup B) -C$  
if $x \in (A \cup B)-c $ then $x \notin C$ this $x \in A$ and/or $x\in B$
Since $x \notin C$ and $x\in (A \cup B)-C$ , $x \in (A-C)$ and/or $x \in (B-C)$
thus if $x \in (A-C)$ and/or $x \in (B-C)$ then $x \in (A-C) \cup (B-C)$
by the definition of union, the same result occurs if x is in only one of both A and B.
Hence, $(A\cup B) - C = (A-C)\cup (B-C)$
