Prove $ (A \cup B) \cap C$ = $(A \cap C) \cup (B \cap C) $

Prove $ (A \cup B) \cap C$ = $(A \cap C) \cup (B \cap C) $

Starting from the left side,

$ (A \cup B) \cap C = $

By distributive law, ( distributing the $\cap C$), we have

$ (A \cap C ) \cup (B \cap C) = $

Therefore,

$ (A \cap C ) \cup (B \cap C) = (A \cap C) \cup (B \cap C)$

If I start from the right, I have

= $(A \cap C) \cup (B \cap C) $

By Distributive Law

= $(A \cup B ) \cap C$

Therefore, $ (A \cup B) \cap C$ = $(A \cup B ) \cap C$

Did I do this correctly or do I need to include the set union definition and the set intersection definitions?

Assuming that I need to include the set union definition of $A \cup B$ for the left side $[x: x \in A \lor x \in B]$ so that means that x belongs in A or x belongs in B

For the right side I would have set intersection

$[x: x \in A \land x \in C]$ so x belongs in A and x belongs in C

$[x: x \in B \land x \in C]$ so x belongs in B and x belongs in C

so maybe it's like this?

$[x: x \in A \land x \in C] \lor [x: x \in B \land x \in C]$

and then by distributive law I would have gotten

$[x: x \in A \lor x \in B] \land C$

which becomes $(A \cup B) \cap C$

My question is how do I write a better proof than this jumbled mess?

$\begin{align*} & & x &\in (A \cup B) \cap C \\ &\iff & x &\in (A \cup B) \wedge x \in C \\ &\iff & (x &\in A \vee x \in B) \wedge x \in C \\ &\iff & (x &\in A \wedge x \in C) \vee (x \in B \wedge x \in C) \\ &\iff & (x &\in A \cap C) \vee (x \in B \cap C) \\ &\iff & x &\in (A \cap C) \cup (B \cap C) \\ \end{align*}$

We've only used the definitions of union and intersection and the distributive law of logic. Since every line is an equivalence, the first and last lines are equivalent.