Set Proof verification 
How to prove that $(A-B)\cup(A\cap B)=A$?

Here is my proof, I'm not sure if I verified it correctly.
$(A-B) \cup (A \cap B)$ simplifies into $x \in (A-B)  \vee x \in (A \cap B)$. 
Then we have $[x \in A$ and $x \notin B$] or $[x \in A$ and $x \in B$] $\iff x \in A$
Therefore, $(A-B) \cup (A \cap B)  = A$  

Provide another proof:
Notice that $(A-B) \subset A$ and $(A\cap B) \subset A$, thus $(A-B) \cup (A\cap B) \subset A$
On the other hand, $\forall x \in A$, if $x\in B$, then $x\in A\cap B$; else if $x\notin B$, then $x\in A-B$. Thus $x\in (A-B) \cup (A\cap B)$. So we have $A\subset (A-B) \cup (A\cap B)$.
So we get $A = (A-B) \cup (A\cap B)$
