Show that
$$\forall x(C \land D) \equiv (\forall x\ C) \lor D$$
where $C$ may have free occurrences of $x$, but $D$ does not have a free occurrence of $x$.
Prove this using only propositional logic inference rules and the quantifier introduction and elimination rules (don't use any of the logical identities).
It is evidently enough to show that (i) from the assumption $\forall x(\varphi(x) \land D)$ you can derive $(\forall x\varphi(x) \land D)$ and (ii) from the assumption $(\forall x\varphi(x) \land D)$ you can derive $\forall x(\varphi(x) \land D)$ (where $D$ doesn't contain $x$ free).
For (i): suppose $\forall x(\varphi(x) \land D)$; instantiate to get $(\varphi(a) \land D)$; use $\land$-elimination to derive $\varphi(a)$ and $D$ separately; use quantifier introduction on the first of those (why is that legitimate)?; and then use $\land$-introduction to get the desired conclusion.
You can now do (ii) similarly.