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?
 A: 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.).

Let me say a bit more about what's going on here.
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!
