Proof of $A\cap(B\cup C) = (A\cap B)\cup(A\cap C)$ I have to resit a calculus exam and for some reason set proofs were never my best friend... Anyway, on a practice exam I encountered the following proof: $$A\cap(B\cup C) = (A\cap B)\cup(A\cap C)$$ When I draw a Venn-diagram it seems quite obvious but I couldn't manage to write the proof down properly. If someone could help me, that'd be great!

If $x\in A\cap(B\cup C)$, then $x\in A$ and $x\in B\cup C$.

$x\in B\cup C\implies (x\in B$ or $x\in C)$.

So, $x\in A\cap(B\cup C)\implies x\in (A\cap B)$ or $ x\in (A\cap C)$

$\implies x\in (A\cap B)\cup(A\cap C)$

$\implies A\cap(B\cup C)\subseteq (A\cap B)\cup(A\cap C)$.

Similarly,

if $y\in (A\cap B)\cup(A\cap C),$

$\implies y\in (A\cap B)$ or $y\in (A\cap C),$

$\implies y\in A$ and $y\in (B$ or $C)$

$\implies y\in A$ and $y\in (B \cup C)$

$\implies y\in A\cap (B \cup C)$.

Now, $A \subseteq B$ and $B \subseteq A \implies A=B$.