Proving $A \cap (B-A) = \emptyset$ (new to proofs!)

Question

I'm a novice mathie - only calculus courses until this past semester, when I started Foundations of Mathematics (a bit of naïve set theory, introduction to basic proofs.)

Up to now, I have been able to complete proofs assigned by using parity, simple modular math, greatest common divisor, etc.

That being said, I am stuck on developing a proof for this.

Let $A$ and $B$ be sets. $A \cap(B-A) = \emptyset$.

The methods I have at my disposal are direct, contrapositive, and contradiction.

I would appreciate a gentle nudge in the right direction. This one has me stumped.

Answer 1

Suppose $x \in A \cap (B-A)$. Then, by property of intersection, $x \in A$ and $\ldots$ what next?

Answer 2

If $x\in B\setminus A$, then $x\notin A$.