How do I do this proof? $(A \cup B) - C = (A - C) \cup B \iff B \cap C = \emptyset$ $(A \cup B) - C = (A - C) \cup B \iff B \cap C = \emptyset$
I know that for this problem, $C$ is removed from $A \cup B$ 
I also know that $(A - C) \cup B$ can be rewritten as $(A \cup B) - (C \cup B)$, but not sure what exactly that means. My assumption is that $A \cup B$ remains after removing $C \cup B$ leaving me with just $A \cup B$, but that doesn't seem to make sense if $C \cup B$ was removed. Am I just left with $A$ in this part?
I'm also a bit confused on how $B \cap C = \emptyset$
Would the proof then go as follows?
1) $(A \cup B) - C \subseteq (A - C) \cup B$
2) $(A - C) \cup B \subseteq (A \cup B) - C$
3) $B \cap C \subseteq \emptyset$
4) $\emptyset \subseteq B \cap C$
Would I then show that:
5) $(A \cup B) - C = (A - C) \cup B \Rightarrow B \cap C = \emptyset$
6) $B \cap C = \emptyset \Rightarrow (A \cup B) - C = (A - C) \cup B$
Am I on the right path? If so, how do I go about formulating each step? Seems like 6 different proofs in one is a bit much.
 A: The first implication:
$$B \cap C  = \emptyset \implies (A \cup B) - C  = (A - C) \cup B $$
Is easy to prove, the problem is the other implication.
You can not prove that something belongs to empty. Therefore, the other demonstration is made by an indirect method.
The statement is equivalent to saying:
$$B \cap C \not = \emptyset \implies (A \cup B) - C \not = (A - C) \cup B $$
Let $x\in(A \cup B)-C \implies x\in(A \cup B) \land x \not \in C
\implies (x\in A \lor x\in B) \land x \not \in C$
$\implies (x\in A \land x\not \in C) \lor (x\in B \land x\not \in C) \implies (x\in(A-C))\lor (x\in (B-C)) $
$\implies x\in((A-C)\cup (B-C))$
But, $B-C \not = B$, because $B \cap C \not = \emptyset$, then $((A-C)\cup B) \not \subseteq ((A-C)\cup (B-C))$, by definition of equality of sets:
$((A-C)\cup (B-C)) \not = ((A-C)\cup B).$ $\blacksquare$
A: Hint:
The following is not right
$$
(A - C) \cup B = (A \cup B) - (C \cup B)
$$
Because
$$
(A - C) \cup B = (A \cup B) - (C -B)
$$
However, we have
$$
(A \cup B) - C= (A-C)\cup (B-C)
$$
Prove that
$$
(A-C)\cup (B-C)=(A - C) \cup B \iff B-C=B
$$
Since $(A-C)\cup (B-C)\subset (A - C) \cup B$),  it only needs to prove
$$
(A - C) \cup B\subset (A-C)\cup (B-C) \iff B-C=B
$$
Then it follows obviously since
$$
B-C=B\iff B\cap C=\varnothing
$$
