Problem: Prove that $\displaystyle (A \cup C) - B = (A-B) \cup C \Leftrightarrow B \cap C = \emptyset $.

Thoughts/Work: I tried proof by contradiction. Suppose that $\displaystyle B \cap C \neq \emptyset $. Then $\displaystyle x \in B $ and $\displaystyle x \in C $. So $\displaystyle x \in (A \cup C) - B \Rightarrow x \in A $ or $\displaystyle x \in C $ and $\displaystyle x \not \in B $. Then $\displaystyle x \in A $ and $\displaystyle x \not \in B $ or $\displaystyle x \in C $. But this contradicts the fact that $\displaystyle B \cap C \neq \emptyset $. So $\displaystyle B \cap C = \emptyset $ for equality between the two sets. Suppose that $\displaystyle (A \cup C) - B \neq (A - B) \cup C $. Then $\displaystyle x \in A $ or $\displaystyle x \in C $ and $\displaystyle x \not \in B \Rightarrow x \not \in A, x \in B $ or $\displaystyle x \not \in C $. So $\displaystyle x \in B $ and $\displaystyle x \not \in C $ or $\displaystyle x \in C $ and $\displaystyle x \not \in B $. This is a contradiction and so $\displaystyle (A \cup C) - B = (A - B) \cup C $. QED

I am not sure if I made this too complicated. Does this look ok?

Thanks