Using only the normal intelim rules, prove $$(\forall x)(\phi (x) \lor G) \equiv [(\forall x)\phi (x) \lor G].$$
For what it's worth, all my attempts have run into trouble over the restriction for $\forall I$ that in moving from $B(T/X)$ to $(\forall X)B$, $T$ must be foreign to all the previous assumptions.
This is an exercise out of an older textbook (for which solutions are not readily available) that's for some reason got me stumped.
Attached is a proof in the Fitch system ... hopefully the rules used correspond to whatever rules you have to work with.