logic and set theory - subset of a set

Question

I want to formally prove that $X\subseteq A\cap B\Leftrightarrow X\subseteq A\wedge X\subseteq B$. This means $\forall x(x\in X\Rightarrow x\in A\wedge x\in B)$. By the use of logic ($P\Rightarrow (Q\wedge R)\equiv (P\Rightarrow Q)\wedge(P\Rightarrow R)$) we get $$ \forall x((x\in X\Rightarrow x\in A)\wedge (x\in X\Rightarrow x\in B)). $$ How can we conclude that this implies $\forall x(x\in X\Rightarrow x\in A)\wedge \forall x(x\in X\Rightarrow x\in B)$ (which is $X\subseteq A\wedge X\subseteq B$)?

Answer 1

$$\forall x (F \wedge G) \vdash \dashv (\forall x (F))\wedge(\forall x(G))$$ So you already have what you need!

Just keep in mind that thiw is not the case with $\vee$ since $$\forall x (F \vee G) \not \vdash (\forall x (F))\vee(\forall x(G))$$