I want to prove that $(A\cap B)\cup(A\cap C) \subseteq (A\cap(B\cup C))$. I noticed that I recently, I have just been applying the laws of logical equivalence (ie: distributivity/commutativity/demorgan's etc). The proof I used was as follows and I wanted to check if this was a valid way because it differed from the solutions.
Proof: Assume $x \in (A\cap B)\cup(A \cap C) \subseteq (A \cap (B \cup C))$
Then $x \in (A \cap B) \vee x \in (A \cap C)$
$ \implies (x \in A \wedge x \in B) \vee (x \in A \wedge x \in C)$
$\implies $ by distributivity of logical statements, $x \in A \wedge (x \in B \vee x \in C)$
Then, $(x \in A) \wedge (x \in B\cup C)$
Thus, $x \in A \cap (B \cup C)$
$\therefore (A\cap B)\cup(A\cap C) \subseteq (A\cap(B\cup C))$
Would this be a valid way as opposed to considering the cases of $x \in (A \cap B)$ and then considering $x \in (A\cap C)$, which was the way the solutions outlined.
Advice would be appreciated.