Why does $\forall n \in \mathbb{N} \vdash P(n)$ not imply $\forall n \in \mathbb{N}P(n)$?

Question

I am trying to understand why $\forall n \in \mathbb{N} \vdash P(n)$ doesn't imply $\forall n \in \mathbb{N}P(n)$ by studying the concepts in this answer, and had 2 questions about the difference between $\forall n \in \mathbb{N} \vdash P(n)$ and $\forall n \in \mathbb{N}P(n)$:

1. Why does $\forall n \in \mathbb{N} \vdash P(n)$ only make sense in the metatheory? I don't understand why the object theory is not capable of talking about whether or not $P(n)$ can be proven given any explicit $n$, especially because the linked answer says that $\vdash \forall n \in \mathbb{N} P(n)$ is a statement made within the object theory. Semantically, why does the quantification in $\forall n \in \mathbb{N} \vdash P(n)$ happen outside of the theory, but the quantification for $\vdash \forall n \in \mathbb{N} P(n)$ happens inside of the theory?
2. Why is the model of the naturals referred to by a statement in the metatheory not necessarily the same as the model of the naturals referred to by a statement in the theory? In this case, both of the statements referring to $\mathbb{N}$, and I thought that $\mathbb{N}$ is a specific model of the naturals, namely $0,1,2,\ldots$; so because we are referring to this set both times, it seems that we are referring to the same model of the naturals regardless of whether we are outside or inside the theory.

For additional context, I am trying to understand why $\forall n \in \mathbb{N} \vdash P(n)$ doesn't imply $\forall n \in \mathbb{N}P(n)$ in order to better understand answers to my question here.

Answer 1

One way of explaining this could be in terms of non-standard models, as in Lumsdaine's comment, which goes as follows:

"if Con(PA) fails in a non-standard model, it means it contains a “proof of non-standard length” of a contradiction from PA. With a little work, one can externalise that to some non-well-founded “proof tree”, which has at each node a possibly-non-standard formula (which in turn externalises to some not-necessarily-well-founded syntax tree), where each step will genuinely follow from earlier steps by some standard rule of proof, but the whole tree will be non-well-founded, and hence not an actual proof."

Thus the proof of negation of Con(PA) can be understood in terms of proofs of nonstandard length and failure of well-foundedness.

Inside a model of $PA + \neg Con(PA)$ one can give a proof of a contradiction from PA; this model and this proof (whether a sequence or a tree of formulas satisfying the deduction rules at each step) are necessarily non-well-founded.

So a metalanguage proof cannot be devised for such a contradiction, but the object (formal) language admits such a proof.

Answer 2

(1) Let's use $\mathbf{n}$ as our numeral for the number $n$. Let $\varphi(x)$ be an open sentence of formal arithmetic. Then of course, if each instance of $\varphi(\mathbf{n})$ is true in the standard model of arithmetic, then $\forall x\varphi(x)$ is true in that model -- since in that model, all the numbers have a corresponding numeral name.

So the crucial first point is this. The claim you are alluding to, about a lack of implication between having available all instances of $\varphi(\mathbf{n})$ and getting $\forall x\varphi(x)$ is a claim about provability, not a claim about truth.

(2) To expand just a little on @spaceisdarkgreen's comment, which makes exactly the key point:

Here's a standard definition:

An arithmetic theory $T$ is $\omega$-incomplete iff, for some open wff $\varphi(x)$, $T$ can prove each wff of the form $\varphi(\textbf{n})$ (where $\textsf{n}$ is $T$'s formal numeral for the number $n$) but $T$ can't go on to prove $\forall x\varphi(x)$.

Look hard at the definition. Certainly there are two different ideas here. Fix on a particular $\varphi(x)$. Then compare claim (A):

$T$ has the deductive power to prove each and every wff of the from $\varphi(\textbf{n}).$

with claim (B):

$T$ has the deductive power to prove the quantified wff $\forall x \varphi(x)$.

And we can very easily imagine an impoverished formal theory $B$ ('Baby Arithmetic') that e.g. proves each wff of the form $\mathbf{2n} = \mathbf{n} + \mathbf{n}$ but can't prove $\forall x(2x = x + x)$ for the boring reason that the formal theory $B$ doesn't know about quantifiers!

(3) OK, but suppose we are dealing with a theory $T$ like Peano Arithmetic which can handle quantified statements. Then, yes, it would of course be nice if whenever (A) is true, then (B) is true. Our guess, pre-1931, might very well be that PA, first-order Peano Arithemtic, is $\omega$-complete.

However, bad news, it follows immediately from the proof of (the syntactic version of) Gödel's first incompleteness theorem that e.g. PA is $\omega$-incomplete. (But to repeat, that's a claim about PA's deductive weakness: of course, for a particular $\varphi$, if each instance of $\varphi(\mathbf{n})$ is true on the standard model, then $\forall x\varphi(x)$ will be true on that model too.)

If you want a proof of that, you'll find an accessible one e.g. on pp. 90-92 of my Gödel Without Too Many Tears which you can download freely from https://www.logicmatters.net/igt. That book will also tell you why we can't beef up PA and make it $\omega$-complete without making it either inconsistent or not computably axiomatizable.