# Prove that $B = (A \cap B) \cup (A' \cap B)$. [duplicate]

I started using the distributive property to open the expression but it goes on and on.

$$(A \cap B) \cup (A' \cap B) = B \cap (A \cup B) \cap (A' \cup B)$$ (Using distributive property)

If I start opening by distributive, the expression gets bigger and bigger, how should I solve it?

• If $b\in B$, either $b\in A$ or $b\notin A$
– JPA
Oct 4, 2020 at 9:46
• Try using the distributive property to factor out $B$ - ie "apply it backwards". Oct 4, 2020 at 9:47

$$B = (A \cup A') \cap B = (A \cap B) \cup (A' \cap B)$$.

Take $$x \in B$$

• If $$x \in A$$, then $$x \in (A \cap B) \implies x \in (A \cap B)\cup (A' \cap B)$$.
• if $$x \notin A$$, then $$x \in A'$$. This implies $$x \in (A' \cap B) \implies x \in (A \cap B)\cup (A' \cap B)$$.

Thus $$B \subseteq (A \cap B)\cup (A' \cap B) \ \ (*)$$

Take $$x \in (A \cap B)\cup (A' \cap B)$$. Then, definitely, $$x \in B$$.

Thus $$(A \cap B)\cup (A' \cap B) \subseteq B \ \ (**)$$

$$(*) \land (**)\implies B = (A \cap B)\cup (A' \cap B) \ \ \blacksquare$$

For a visual explanation, this venn diagram should make it clear:

$$B$$ is the union of the two highlighted areas, $$A \cap B$$ and $$A' \cap B$$. Therefore, $$B = (A \cap B) \cup (A' \cap B)$$.