How do I solve $\overline{(A \cap B) \cup (\overline{A} \cap C)} = (A \cap \overline{B}) \cup (\overline{A} \cap \overline{C})$? The question is to prove  $\overline{(A \cap B) \cup (\overline{A} \cap C)} = (A \cap \overline{B}) \cup (\overline{A} \cap \overline{C})$ using laws of set theory.
I got stuck after the following steps:
$$
\overline{(A \cap B) \cup (\overline{A} \cap C)} = \overline{(A \cap B)} \cap \overline{(\overline{A} \cap C)}
\\ = (\overline{A} \cup \overline{B}) \cap (A \cup \overline{C})
\\ = ((\overline{A} \cup \overline{B}) \cap A) \cup ((\overline{A} \cup \overline{B}) \cap \overline{C})
\\ = (\overline{A} \cap A) \cup (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C})
\\ = (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C})
$$
Not sure what to do next. I somehow feel that the absorption law could be used here, but don't know how.
 A: Taking what you have done so far
$$\begin{align}
\overline{(A \cap B) \cup (\overline{A} \cap C)} & = \overline{(A \cap B)} \cap \overline{(\overline{A} \cap C)}\\
& = (\overline{A} \cup \overline{B}) \cap (A \cup \overline{C})\\
& = ((\overline{A} \cup \overline{B}) \cap A) \cup ((\overline{A} \cup \overline{B}) \cap \overline{C})\\
& = (\overline{A} \cap A) \cup (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C})\\
& = (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C}) & \tag{1}
\end{align}$$
it looks fine.
Next you go as follows. We know from definition of union that
$$(\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \subseteq (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C})$$
Although
$$\begin{align}
x \in (\overline{B} \cap \overline {C}) & \implies x \in \overline{B} \wedge x \in \overline{C}\\
& \implies \big(x \in \overline{B} \wedge x \in \overline{C} \big) \wedge(\overline{A} \vee A)\\
& \implies \big(x \in \overline{B} \wedge x \in \overline{C} \wedge x \in \overline{A} \big) \vee \big( x \in \overline{B} \wedge x \in \overline{C} \wedge x \in A \big)\\
& \implies (x \in \overline{C} \wedge x \in \overline{A}) \vee (x \in \overline{B} \wedge x \in A)\\
& \implies (x \in \overline{C} \cap \overline{A}) \vee (x \in \overline{B}\cap A)\\
& \implies x \in (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C})
\end{align}$$
which means that $(\overline{B} \cap \overline{C}) \subseteq (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}).$ So we conclude that
$$\begin{align}
(\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C}) & \subseteq (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C})\\
& \subseteq (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C})
\end{align}$$
Continuing from $(1)$ we get the following final conclusion
$$\therefore (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C}) = (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \quad \square$$
A: Hint: To show that $$ (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C})$$ is same as $$ (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) $$ it is enough to show that $$(\overline{B} \cap \overline{C})$$ is  a subset of $$ (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C})$$ Prove this by writing $E$ as $(E \cap A) \cup (E \cap \overline A)$ where $E$ stands for $$(\overline{B} \cap \overline{C})$$
