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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?

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

3 Answers 3

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

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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 \ \ (**)$


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

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

venn diagram

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

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