# Necessity of universal quantifier in predicate calculus

I am reading Kleene's "Introduction to Metamathematics". There in Chapter 7 Section 32 he mentions two interpretations of free variables in predicate calculus.

One of them is that "For the generality interpretation of a variable $$x$$ in $$A(x)$$ the context within which all free occurrences of $$x$$ must represent the same object is exactly the whole $$A(x)$$. The formula $$A(x)$$ then means the same as $$\forall x A(x)$$."

Question: If $$A(x)$$ means the same as $$\forall x A(x)$$, then why do we introduce $$\forall$$ at all?

Consider the sentences $$\forall x(P(x)\vee Q(x))$$ and $$(\forall x(P(x)))\vee (\forall x(Q(x))).$$ If we remove the $$\forall$$s, these can't be distinguished, but they obviously should be (take $$P$$ = "is even" and $$Q$$ = "is odd," e.g.).
Kleene is essentially describing an operation on formulas - universal closure. Given a formula $$\varphi$$, the universal closure of $$\varphi$$ is gotten by universally quantifying over all the free variables of $$\varphi$$. E.g. the universal closure of $$P(x)\wedge\exists y(Q(y,z))$$ would be $$\forall x\forall z[P(x)\wedge\exists y(Q(y,z))]$$. The universal closure operation ignores all-binding universal quantifiers, but it doesn't ignore universal quantifiers in general.
So why does this feel weird? Well, intuitively we want equivalence to be "structural" - if $$\varphi$$ is equivalent to $$\hat{\varphi}$$ then we should be able to take a formula $$\psi$$ and replace all $$\varphi$$s in $$\psi$$ with $$\hat{\psi}$$s without changing the meaning of $$\psi$$. The equivalence between $$P(x)$$ and $$\forall x(P(x))$$ gotten when we adopt the semantics you've described above, however, is not structural in this sense. Incidentally, I take this as a reason to not do that!