Prove the following $(A \cap B) \cap (\neg (A\cap C))=A\cap ( B \cap(\neg C))$ I want to prove the following statement
$$(A \cap B) \cap (\neg (A\cap C))=A\cap ( B \cap(\neg C))$$
Rewrite the LHS and RHS:
$$(A\setminus \neg B)\setminus (A \setminus \neg C)=A\setminus (\neg(B\setminus C))$$
RHS can be written as $$(A\setminus \neg B)\setminus (A \setminus \neg C)= A\setminus (\neg B\setminus \neg C)$$ so I get:$$A\setminus (\neg B\setminus \neg C)=A\setminus (\neg(B\setminus C))$$
if I open the new RHS it will be equal to the new LHS?
thanks.

Just apply one of the De Morgan laws and distributivity to the original lefthand side:
$$\begin{align*}
(A\cap B)\cap\neg(A\cap C)&=A\cap B\cap(\neg A\cup\neg C)\\
&=(A\cap B\cap\neg A)\cup(A\cap B\cap\neg C)\\
&=\varnothing\cup(A\cap B\cap\neg C)\\
&=A\cap B\cap\neg C
\end{align*}$$
