TITLE: Prove distributive law of sets
QUESTION [9 upvotes]: Let $A,B,C$ be sets. Prove the distributive law
$$A \cap\left(B\cup C\right) = \left(A\cap B\right)\cup \left(A\cap C\right)$$

proof
First we'll show that $A \cap\left(B\cup C\right) \subset \left(A\cap B\right)\cup \left(A\cap C\right)$, and then the converse. 
If $x$ is in $A \cap\left(B\cup C\right)$, then $x$ must be in $A$ and $x$ must be in $B$ or $C$. An element $x$ can satisfy this membership by being in either $A$ and $B$, or $A$ and $C$. In symbols, 
$$x \in \left[\left(A\cap B\right)\cup \left(A\cap C\right)\right]$$

I'm not going to write the other direction. Because I know once I have this way down I can go the other way. What I'm looking for is -- in addition to correctness -- whether my argument is clunky or ambiguous. This "or" and "and" wording throws me for a loop when I'm trying to use it as a math operation and a word. Ideally, I'd like to know if there's a simpler way to write these types of proofs.

REPLY [2 votes]: I also struggled with the usefulness of spelling out the operations in plain English in this very proof, I feel they don't really add to the proof any more than the symbolic equations. I think you may be able to directly write:
$$
\forall x (x \in (A \cap (B \cup C)) \implies x \in ((A \cap B) \cup (B \cap C))) \implies (A \cap (B \cup C)) \subseteq  ((A \cap B) \cup (B \cap C)))
$$
$$
\forall x (x \in ((A \cap B) \cup (B \cap C)) \implies  x \in (A \cap (B \cup C)) ) \implies ((A \cap B) \cup (B \cap C))  \subseteq  (A \cap (B \cup C)) )
$$
Therefore
$$
(A \cap (B \cup C)) = ((A \cap B) \cup (B \cap C))
$$
The proof relies on only two things: (1) definition of a subset $
A \subseteq B \iff \forall x (x \in A \implies x \in B)
$ and (2) $(X \subseteq Y) \land (Y \subseteq X) \implies X = Y$