# Union with Intersection equals Intersection with Union iff Subset

## Theorem

Let $A$, $B$ and $C$ be sets.

Then:

$\paren {A \cap B} \cup C = A \cap \paren {B \cup C} \iff C \subseteq A$

## Proof

### Necessary Condition

Let $C \subseteq A$.

We have:

 $\displaystyle C$ $\subseteq$ $\displaystyle A$ $\displaystyle \leadsto \ \$ $\displaystyle A \cap C$ $=$ $\displaystyle C$ Intersection with Subset is Subset‎ $\displaystyle \leadsto \ \$ $\displaystyle \paren {A \cap B} \cup \paren {A \cap C}$ $=$ $\displaystyle \paren {A \cap B} \cup C$ $\displaystyle \leadsto \ \$ $\displaystyle A \cap \paren {B \cup C}$ $=$ $\displaystyle \paren {A \cap B} \cup C$ Intersection Distributes over Union

and another way:

 $\displaystyle C$ $\subseteq$ $\displaystyle A$ $\displaystyle \leadsto \ \$ $\displaystyle A \cup C$ $=$ $\displaystyle A$ Union with Superset is Superset‎‎ $\displaystyle \leadsto \ \$ $\displaystyle \paren {A \cup C} \cap \paren {B \cup C}$ $=$ $\displaystyle A \cap \paren {B \cup C}$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {A \cap B} \cup C$ $=$ $\displaystyle A \cap \paren {B \cup C}$ Intersection Distributes over Union

Thus:

$C \subseteq A \implies \paren {A \cap B} \cup C = A \cap \paren {B \cup C}$

$\Box$

### Sufficient Condition

Let $\paren {A \cap B} \cup C = A \cap \paren {B \cup C}$.

Aiming for a contradiction, suppose $\exists x \in C: x \notin A$.

$C \subseteq \paren {A \cap B} \cup C$

and so by definition of subset:

$x \in \paren {A \cap B} \cup C$
$A \cap \paren {B \cup C} \subseteq A$

and so (indirectly) by definition of subset:

$x \notin A \cap \paren {B \cup C}$

Hence by definition of subset:

$\paren {A \cap B} \cup C \nsubseteq A \cap \paren {B \cup C}$

that is:

$\paren {A \cap B} \cup C \ne A \cap \paren {B \cup C}$

$\neg \exists x \in C: x \notin A$
$\forall x \in C: x \in A$
and so $C \subseteq A$ by definition of subset.
$\blacksquare$