*This question misses a point in Artemov's paper, namely that the relevant property of the p.r. function in (2) should itself be PA-provable; see Definition 1 on page 9. This leads to a less trivial situation, and this is what my answer addresses:*

Suppose $\varphi(x)$ is a $\Delta_0$ formula such that $\mathbb{N}\models\forall x\varphi(x)$. I claim the following are equivalent:

There is a primitive recursive function $g$ such that $\mathsf{PA}$ proves "for every $n$, if $\varphi(n)$ fails then $g(n)$ is a $\mathsf{PA}$-proof of $\perp$." (Note that this is *prima facie* stronger than $\mathsf{PA}+Con(\mathsf{PA})\vdash\forall x\varphi(x)$.)

There is some primitive recursive $f$ such that $\mathsf{PA}$ proves "for each $n$, $f(n)$ is a $\mathsf{PA}$-proof of $\varphi(n)$."

$1\rightarrow 2$: Fix $g$ as in $1$, and consider the function $f$ defined as follows. On input $n$ we first check whether $\varphi(n)$ is actually true; the "brute force verification" proof of this fact, if indeed it is true, can be generated primitive recursively. If $\varphi(n)$ holds we output this brute force verificatino. If on the other hand we see that $\varphi(n)$ is false, then we output $g(n)$. This can all be bundled together as a primitive recursive function (the key point being that verifying/falsifying a $\Delta_0$ sentence can be done quickly), so we have $2$.

$2\rightarrow 1$: Fix $f$ as in $2$, and consider the function $g$ defined as follows. On input $n$ we first check whether $\varphi(n)$ holds. If it does, we output $17$ (or whatever). If $\varphi(n)$ fails, then we output the $\mathsf{PA}$-proof of $\perp$ gotten by combining the (short) falsification of $\varphi(n)$ with the proof $f(n)$.

So what? Well, this tells us that if $\mathsf{PA}$ "Artemov-proves" $Con(T)$, then we already have $\mathsf{PA}+Con(\mathsf{PA})\vdash Con(T)$. So in a reasonably strong sense we get that $\mathsf{PA}$ can't go "past itself" in Artemov's sense.

As a digression, condition 1 does suggest something interesting. Given $\Pi^0_1$ sentences $\alpha\equiv\forall x\psi(x)$ and $\beta\equiv\forall x\theta(x)$ (with $\psi,\theta\in\Delta_0$), write "$\alpha\le_\mathsf{PA}\beta$" iff there is a primitive recursive function $f$ such that $\mathsf{PA}\vdash\forall n[\neg\psi(n)\rightarrow\neg \theta(f(n))]$. *(Since $\Delta_0$ statements can be efficiently checked I don't see any value in shifting from instances to proofs here.)*

This gives rise to a degree structure on the $\Pi^0_1$ sentences. The false $\Pi^0_1$ statements are appropriately maximal, while the argument above (suitably tweaked) shows that $Con(\mathsf{PA})$ is the $\le_\mathsf{PA}$-maximal element of the set of "Artemov-provable" sentences. I strongly suspect this degree structure is well-known in proof theory, but I'm not a proof theorist. Any references welcome!

*Note that if we replaced "primitive recursive" with "PA-provably-total" this would just be the $\Pi^0_1$ part of the Lindenbaum algebra over $\mathsf{PA}$: if $\mathsf{PA}\vdash\beta\rightarrow\alpha$ then $\mathsf{PA}$ proves that the function sending each counterexample to $\alpha$ to the smallest counterexample to $\beta$, and sending everything else to $17$, is total.*

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