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

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, $(A\cup B) - C = (A-C)\cup (B-C)$
