proof by contradiction $A ∩ (B - A)= \varnothing.$ Use the method of proof by contradiction to prove that if A and B are sets, then
$$A ∩ (B - A)= \varnothing.$$
It says I have to use contradiction, but contradiction is the one I have a problem with!
 A: To prove a statement by contradiction, you assume the negation of what it is you need to prove, and then work to obtain a contradiction.
So to prove that $(P \land Q \land R) \rightarrow S$, you assume $P \land Q \land R$, along with the negation of $S$: $\lnot S$, and then show that this assumption of $\lnot S$ contradicts one or more of our givens: $P$ or $Q$ or $R$, or that it contradicts any other fact or axiom we know to be true.
So in this problem, we take as given that $A$ and $B$ are sets, and we assume the negation of $A ∩ (B – A)= ∅.$ This would mean we assume that $A \cap (B - A) \neq \varnothing$. 
So we assume for the sake of contradiction that there exists at least one $x \in A \cap (A -B)$. And thus it would follow that $$\begin{align} x \in A\cap (A - B) & \iff x \in A\;\text{ and }\;x \in (B - A) \\ \\ &\iff x \in A \;\text{ and }\; (x\in B \land x\notin A) \\ \\ & \iff (x \in A \;\text{and}\; x\in B\;\text{and}\; x \notin A)\end{align}$$
Can you see that we've reached a contradiction?
This means our assumption that $A \cap (B - A) \neq \varnothing$ is FALSE, and so it must be true that $A \cap (B - A) = \varnothing$.
A: Start with this. 
"Suppose that $A \cap (B - A) \ne \emptyset$. Then there's an element $x \in A \cap (B - A)$. That implies that (1) $x \in A$ and (2) ..."
Then keep writing until you get two statements that seem to contradict one another. 
A: Suppose by contradiction that there is an element $$x\in A\cap (B-A).$$ Note that $$A\cap(B-A)=A\cap(B\cap A^c)=B\cap (A\cap A^c)$$ i.e. $x\in B$ and $x\in \emptyset$. ($\Rightarrow \Leftarrow)$
