Union with Intersection equals Intersection with Union iff Subset

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

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

Necessary Condition
Let $C \subseteq A$.

We have:

and another way:

Thus:


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

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

$\exists x \in C: x \notin A$.

From Set is Subset of Union:
 * $C \subseteq \paren {A \cap B} \cup C$

and so by definition of subset:
 * $x \in \paren {A \cap B} \cup C$

From Intersection is Subset:
 * $A \cap \paren {B \cup C} \subseteq A$

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

This contradicts our initial assertion.

Hence by Proof by Contradiction:
 * $\neg \exists x \in C: x \notin A$

From Assertion of Universality:
 * $\forall x \in C: x \in A$

and so $C \subseteq A$ by definition of subset.