set theory proof of $A\cap B = A \setminus(A\setminus B)$ 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.

Notice that
$$\begin{align}A \setminus (A \setminus B) & = A \cap (A \cap B^c)^c = A \cap (A^c \cup B) \\ & = (A \cap A^c) \cup (A \cap B) = \varnothing \cup (A \cap B) = A \cap B.\end{align}$$
I have used two properties here - the de Morgan's laws for sets and the distributivity of $\cup$ and $\cap$.
