Prove that in set theory $A-B = A - (A \cap B)$ Prove that in set theory $A-B = A - (A \cap B)$
Please give me a hint.
Let $x\in A-B \implies x\in A~ and ~x \notin B\implies x\in A ~and (x\in A ~and~ x \notin B)\implies x\in A - (A \cap B)$

Use the definition of $A-B$, namely $A\cap \overline{B}$, the De Morgan's laws, the Distributive laws, the Complement laws and the Identity laws.
$$A-(A\cap B)=A\cap\overline{A\cap B}$$
$$=A\cap(\overline{A}\cup \overline{B})=(A\cap \overline{A})\cup(A\cap\overline{B})$$
$$=\varnothing\cup(A\cap \overline{B})=A\cap \overline{B}$$
