Prove $A \subseteq (B \cap C) \iff (A \subseteq B)$ and $(A \subseteq C)$ I need some help with the following proof:

$A \subseteq (B \cap C) \iff (A \subseteq B) \text{ and }(A \subseteq C)$.

Any help would be appreciated.
 A: First: I suggest you convince yourself that $B\cap C \subseteq B$ and $B\cap C \subseteq C$, using only the definition of the intersection of sets, and of a subset of a set.
You can use element chasing to prove this, too.
$$[x \in B\cap C \iff (x \in B \text{ and } x \in C)] \iff (B\cap C \subseteq B \text{ and } B\cap C\subseteq C)$$
Now, you can use the above, and the method of element chasing, to prove $$A \subseteq (B \cap C) \iff (A \subseteq B) \text{ and }(A \subseteq C)\tag{$\dagger$}$$
recalling that $$X \subseteq Y \iff (x \in X \rightarrow x \in Y)$$
You can prove bidirectionality $(\iff)$ of the claim to be proven ($\dagger$) by sticking strictly to definitions.
A: As a comment says, just use the definitions, and use the laws of logic...  You can calculate
\begin{align}
& A \subseteq B \cap C \\
= & \qquad \text{"definition of $\;\subseteq\;$"} \\
& \langle \forall x :: x \in A \implies x \in B \cap C \rangle \\
= & \qquad \text{"definition of $\;\cap\;$"} \\
& \langle \forall x :: x \in A \implies x \in B \land x \in C \rangle \\
= & \qquad \text{"logic: ..."} \\
\end{align}
Now do you see how to use two distribution rules from logic to get two separate parts, one about $\;B\;$ and one about $\;C\;$, so that you end up with the equivalent of $\;(A \subseteq B) \land (A \subseteq C)\;$?
You might want to rewrite $\;P \implies Q\;$ to $\;\lnot P \lor Q\;$, if that makes it easier.
