Prove for all sets A,B,C: If $C-B=\varnothing$ then $(A\cup C)-(B-C^c)=A-(B\cap C)$ To be proven: If $C-B=\varnothing$ then $(A\cup C)-(B-C^c)=A-(B\cap C)$.
I've been stuck on this problem for a few days, I try to use set identities on both sides of the equation to try and make them equal but I can't figure it out.
Here's where I keep getting stuck
$$(A\cup C)-(B-C^c) = (A\cup C)-(B\cap(C^c)^c) = (A\cup C)-(B\cap C)$$
I don't know how to make $(A\cup C)$ into $A$ so that the equation is true.
 A: For any two sets you have $$X-Y=X\cap Y^c.$$
Using this you get $B-C^c=B\cap C$ and
$$(A\cup C) - (B-C^c)=\\
(A\cup C) - (B\cap C)=\\
(A\cup C) \cap (B\cap C)^c = \\
(A\cup C) \cap (B^c\cup C^c) =\\
A\cap (B^c\cup C^c) \cup C \cap (B^c\cup C^c) =\\
A\cap (B^c\cup C^c) \cup \emptyset =\\
A\cap (B\cap C)^c =\\
A- (B\cap C)
$$
A: Here's a graphic proof.

$C-B$ is empty, so we draw x's there.
Let region 1 = $A \cup C$, 2 = $B - C^c$, 3 = $A$ and 4 = $B \cap C$. As you can see, $1 - 2 = 3 - 4$. 
Even if you may ultimately want an algebraic proof, drawing a Venn diagram may indicate how that proof should go.
A: If:$$R:=A-\left(B\cap C\right)=A\cap\left(B\cap C\right)^{c}$$ Then:
$$L:=\left(A\cup C\right)-\left(B-C^{c}\right)=\left(A\cup C\right)\cap\left(B-C^{c}\right)^{c}=\left(A\cup C\right)\cap\left(B\cap C\right)^{c}=$$$$\left[A\cap\left(B\cap C\right)^{c}\right]\cup\left[C\cap\left(B\cap C\right)^{c}\right]=R\cup\left[C\cap\left(B\cap C\right)^{c}\right]$$
This with:
$$C\cap\left(B\cap C\right)^{c}=C\cap\left(B^{c}\cup C^{c}\right)=\left(C\cap B^{c}\right)\cup\left(C\cap C^{c}\right)=\varnothing\cup\varnothing=\varnothing$$
So: $$L=R$$
