How can you prove that: A-(A∩B)=A-B ?

Question

I know this might be stupid for most of the people here. I am starting with set theory and I cannot prove this:A-(A∩B)=A-B

I start by assuming: 1) x∈A & ¬ (x∈A & x∈B)

     Therefore I understand I can infer: 

             2) x∈A                  from 1
             3) ¬ (x∈A & x∈B)        from 1
             4) ¬x∈A & ¬x∈A          from 3
             5) ¬x∈A                 from 4
             6) ⊥                    from 2 and 5

I do not know how to turn around this contradiction. At an intuitively level it is very clear to me, but I am not able to formalize it.

Answer 1

You'd do this by set inclusion. If we can show that $A - (A \cap B) \subset A-B$, and that $A-B \subset A - (A \cap B)$, we'd conclude equality.

So let $x \in A-B$. Then $x \in A, x \notin B$, so $x \notin A \cap B$, and so $x \in A - (A \cap B)$.

If $x \in A - (A \cap B)$, $x\in A$ but not in $ A \cap B$. If $x \in B$ we'd have a contradiction. That finishes the proof.

Answer 2

\begin{align*} x \in A - (A\cap B) &\iff x \in A \; \wedge \; x \notin A\cap B \\ &\iff x \in A \; \wedge \; (x\notin A \: \vee \: x \notin B) \\ &\iff x \in A \; \wedge \; x \notin B \\ &\iff x \in A -B. \end{align*}