Proof of set identities The question asks to prove that   $$(A\cup B')\cap(A'\cup B) = (A\cap B) \cup (A'\cap B')$$ where $A,B$ are sets. How could could i approach and solve this question, and also if there are additional resources that could help me with these kind of problems?
 A: By the distributive law
$$(A\cup B')\cap(A'\cup B)=((A\cup B')\cap A')\cup((A\cap B') \cap B)\overset{(*)}=(B'\cap A')\cup(A\cap B)$$ since $(*)$ $A'\cap A=\emptyset$ and $B'\cap B=\emptyset$. Now, $$(B'\cap A')=(A'\cap B')\quad \text{ and }\quad (B'\cap A')\cup(A\cap B)=(A\cap B)\cup (B'\cap A')$$ by the commutative law (see the same link).

You can find more details about the rules that apply in the algebra of sets in the link above.
A: $$
x\in(A\cup B')\cap(A'\cup B)\\
\Updownarrow\\
x\in(A\cup B')\wedge x\in(A'\cup B)\\
\Updownarrow\\
[x\in A\vee x\in B']\wedge[x\in A'\vee x\in B]\\
\Updownarrow(\star)\\
[x\in A\,\wedge[x\in A'\vee x\in B]]\vee[x\in B'\wedge[x\in A'\vee x\in B]]\\
\Updownarrow(\star\star)\\
[x\in A\,\wedge\,x\in A']\vee[x\in A\,\wedge\,x\in B]\vee[x\in B'\,\wedge\,x\in A']\vee[x\in B'\,\wedge\,x\in B]\\
\Updownarrow(\star\star\star)\\
[x\in A\,\wedge\,x\in B]\vee[x\in B'\,\wedge\,x\in A']\\
\Updownarrow(\star\star\star\,\star)\\
[x\in A\,\wedge\,x\in B]\vee[x\in A'\,\wedge\,x\in B']\\
\Updownarrow\\
x\in(A\cap B)\cup(A'\cap B')
$$
where $(\star)$ follows from the distributivity of conjunction over disjunction in mathematical logic, $(\star\star)$ also follows from the distributivity of conjunction over disjunction and from the associativity of disjunction in mathematical logic (we skipped one step), $(\star\star\star)$ follows from the fact that $[x\in A\,\wedge\,x\in A']$ and $[x\in B'\,\wedge\,x\in B]$ are always false (by definition of $A'$ and $B'$) and $(\star\star\star\,\star)$ follows from the commutativity of conjunction in mathematical logic.
