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.

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.
