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.

For any two sets you have $$X-Y=X\cap Y^c.$$
Using this you get $B-C^c=B\cap C$ and
$$(A\cup C) - (B-C^c)=\\
(A\cup C) - (B\cap C)=\\
(A\cup C) \cap (B\cap C)^c = \\
(A\cup C) \cap (B^c\cup C^c) =\\
A\cap (B^c\cup C^c) \cup C \cap (B^c\cup C^c) =\\
A\cap (B^c\cup C^c) \cup \emptyset =\\
A\cap (B\cap C)^c =\\
A- (B\cap C)
$$
