set theory proof of $A\cap B = A \setminus(A\setminus B)$ [duplicate]

Please proof that $A\cap B=A\setminus(A\setminus B)$ this really gave me sleepless night. I tried using the set intersection properties but I still got confused.

To show $A\cap B = A -(A-B)$, you show two subsets relations $A\cap B \supseteq A -(A-B)$ and $A\cap B \subseteq A -(A-B)$.

To show $A\cap B \supseteq A -(A-B)$, let $x \in A-(A-B)$, then $x\in A$ and $x\notin A-B$. Here if $x\notin B$, then $x\in A-B$ and contradiction arises, so $x\in B$. so $x\in A$ and $x\in B$, i.e. $x\in A\cap B$. So we show $A\cap B \supseteq A -(A-B)$.

To show $A\cap B \subseteq A -(A-B)$, let $x\in A\cap B$, then $x\in A$. Here if $x\in A-B$, then $x\notin B$ and contradiction arises, so $x\notin A-B$. Together we have $x\in A$ and $x\notin A-B$, i.e. $x\in A-(A-B)$. So $A\cap B \subseteq A -(A-B)$. Hence this completes the proof.