Show that $(A\cap A')\setminus (B\cap B')=(A\setminus B)\cap (A'\setminus B')$ I want to show that $$(A\cap A')\setminus (B\cap B')=(A\setminus B)\cap (A'\setminus B')$$ for arbitrary sets $A,B,A',B'$.
I have done the following:
Let $x\in (A\cap A')\setminus (B\cap B')$.
Then we have that
\begin{align*}x\in &(A\cap A')\setminus (B\cap B') \iff x\in (A\cap A') \land x\notin  (B\cap B') \\ &\iff ((x\in A)\land (x\in A')) \land x\in  (B\cap B')^c \\ &\iff ((x\in A)\land (x\in A')) \land x\in  (B^c\cup B'^c) \\ &\iff ((x\in A)\land (x\in A')) \land   ((x\in B^c)\lor (x\in B'^c)) \\ &\iff ((x\in A)\land (x\in A')\land (x\in B^c))    \lor ((x\in A)\land (x\in A') \land (x\in B'^c)) \end{align*}
Is everything correct so far? How could we continue?
 A: I believe the equality that you mentioned
$$(A\cap A')\setminus (B\cap B')=(A\setminus B)\cap (A'\setminus B') \tag{1}$$
is not true, in general.
Consider the following.
$$\begin{align}
x \in (A \cap A’) \setminus (B \cap B’) & \iff x \in A \cap A’ \wedge x \notin B \cap B’\\
& \iff \big(x \in A \wedge x \in A’\big) \wedge \big( x \notin B \vee x \notin B’ \big)\\
& \iff \big( x \in A \wedge x \in A’ \wedge x \notin B \big)\\
& \quad \quad \quad \vee \big(x \in A \wedge x \in A’ \wedge x \notin B’\big)\\
& \implies \big( x \in A \wedge x \notin B \big) \vee \big( x \in A’ \wedge x \notin B’ \big)\\
& \iff (x \in A \setminus B) \vee \big( x \in A’ \setminus B’ \big)\\
& \iff x \in (A \setminus B) \cup (A’ \setminus B’)
\end{align}$$
From this I can just get a counter example that shows that $(1)$ doesn’t hold.
$$A = \{1,2\} \quad A’ = \{1\} \quad B = \{1\} \quad B’ = \{2\}$$
$$\begin{align}
1 \in A,A’,B \wedge 1 \notin B’ & \implies 1 \in A \cap A’ \wedge 1 \notin B \cap B’\\
& \implies 1 \in (A \cap A’) \setminus (B \cap B’)
\end{align}$$
$$1 \in A,B \implies 1 \notin A \setminus B \implies 1 \notin (A\setminus B) \cap (A’ \setminus B’)$$
A: A correct formula would be the following:
$$ (A\smallsetminus B)\cap(A'\smallsetminus B')=(A\cap A')\smallsetminus(B\cup B'), $$
which is easy to prove:
\begin{align}
(A\smallsetminus B)\cap(A'\smallsetminus B')&=(A\cap B^c)\cap(A'\cap B'^c) \\
&=(A\cap A')\cap (B^c\cap B'^c)\\
&=(A\cap A')\cap (B\cup B')^c=(A\cap A')\smallsetminus(B\cup B').
\end{align}
