# prove $A \cap (B - A) = \emptyset$

I'm trying to prove using the contradiction method that $$A \cap (B - A) = \emptyset.$$

Since I've to prove it using the contradiction method, we assume $A \cap (B - A) = \{x_1,x_2,\dots,x_n\} \ne \emptyset$ Then, I tried creating two different scenarios:

1) There's no common elements between $A$ and $B$ so $(B-A) = B$, therefore $A \cap (B-A) = \emptyset$

2) There are common elements between $A$ and $B$

In the second scenario is the one I'm stuck which should give a contradiction showing that it will end on an empty set.

we can rewrite it a bit and possibly make things clearer:$$(B-A)=B\setminus A= \{b:b\in B, b\notin A\}$$ then:$$A\cap (B-A)$$ becomes$$\{b:b\in A \land b\in B \land b\notin A\}$$ but the first of these, and the last, logically contradict. b can't both be in A, and not in A ( at least in finite sets). this is set builder notation, and $\land$ is the logical AND.
$$A \cap (B-A)= A \cap (B \cap A^C)= B \cap (A \cap A^C) = B \cap \emptyset = \emptyset$$
Note that an element cannot be in $B$ and in the complement of $B$ at the same time.
Namely we cannot $x \in B$ and $x \notin B$
If $x \in A \cap( B$ \ $A)$ then $x \in A$ and $x \in B$ \ $A=\{x \in B \wedge x \notin A\} \Rightarrow x \in A \wedge x \notin A$ which is a contradiction.