Closure Theorem (Shoenfield Mathematematical Logic)

Question

This theorem states the following : Let A be a formula and A' be its closure(all the free variables of A are quantified in A'). Then A is provable iff A' is provable. I understand that when the closure of the formula contains just existential quantifiers, but why is it true in the case where A' contains universal quantifiers?

Answer 1

It seems that your statement of the theorem is missing a word. $A'$ should be the universal closure of $A$ -- that is, every free variable in $A$ is unversally quantified to produce $A'$.

The theorem is not true if existential quantifiers can also be used. For example, in a pure predicate calculus (with equality) in a logical language that contains a constant symbol $0$, the formula $x=0$ is not provable, but $\exists x(x=0)$ is.