Let $M$ be a nonstandard model of $PA$.



> 
> Q1. Is there any way to get a submodel $N\subset M$ such that $N\models Th\_{\Pi\_2}(PA)$, but $N\not\models PA$?
> 
> 
> Q2. Especially, what combinatorial principle can be used to construct a model $N\models Th\_{\Pi\_2}(PA)$ such that $N\not\models PA$? Or $Th\_{\Pi\_2}(PA)\vdash PA$?
> 
> 
> 



On models of $Th_{\Pi_2}(PA)$
$\def\pa{\mathit{PA}}\def\rfn{\mathrm{RFN}}\def\pr{\mathrm{Pr}}\def\num#1{\ulcorner#1\urcorner}\def\Th{\mathrm{Th}}$ $\Th\_{\Pi\_2}(\pa)$ certainly does not prove $\pa$:



> 
> **Theorem:** For any constant $k$, there is no consistent $\Pi\_k$-axiomatized theory $T$ such that $T\vdash\pa$.
> 
> 
> More generally, if $S$ is a [sequential theory that proves full induction](https://mathoverflow.net/a/87249), then $S$ is not derivable from any consistent theory $T$ axiomatized by formulas of restricted quantifier complexity.
> 
> 
> 


(So, for example, $\mathit{ZF}$ is not derivable from any consistent theory axiomatized by $\Pi\_k$-sentences in the Levy hierarchy.)


The reason is that $\pa$ proves the uniform reflection principles
$$\rfn\_Q(\Pi\_k)=\forall x\,(\pr\_Q(x)\land x\in\Pi\_k\to\mathrm{Tr}\_k(x)),$$
where $\mathrm{Tr}\_k$ is the truth predicate for $\Pi\_k$-sentences, $\mathrm{Pr}\_Q$ is the provability predicate for $Q$, and $Q$ could really stand for any fixed finitely axiomatized subtheory of $\pa$ (the choice does not matter much).


So, if $T$ is a $\Pi\_k$-axiomatized theory that proves $\pa$, then there is a finitely axiomatized subtheory $T\_0\subseteq T$ that proves $\rfn\_Q(\Pi\_{k+1})$, and (say) $I\Sigma\_1$. Let $\psi=\bigwedge T\_0$. Then
$$\pr\_Q(\num{\neg\psi})\to\neg\psi$$
is an instance of $\rfn\_Q(\Pi\_{k+1})$, which implies that $T\_0$ proves its own consistency. By Gödel’s theorem, $T\_0$ is inconsistent, hence so is $T$.


As for constructing submodels of $M\models\pa$ satisfying $\Th\_{\Pi\_2}(\pa)$ but not $\pa$, there is the following general result due to McAloon [1]:



> 
> **Theorem (McAloon):** Let $T$ be a recursively axiomatizable $\Sigma\_1$-sound extension of $I\Sigma\_1$. Then for any countable nonstandard model $M\models I\Delta\_0$, there is a nonstandard cut $N\subseteq M$ such that $N\models T$.
> 
> 
> 


This was generalized to $M\models IE\_1$ by Wilmers [2].


McAloon’s argument relies on the indicator theory, hence one may consider it a combinatorial construction.


In order to get $N\models\Th\_{\Pi\_2}(\pa)$, $N\not\models\pa$, we can apply this theorem to
$$T=I\Sigma\_1+\Th\_{\Pi\_2}(\pa)+\neg\rfn\_Q(\Pi\_4).$$
We only need to check that $T$ is $\Sigma\_1$-sound. In fact, $T$ is $\Pi\_4$-conservative over the (sound) theory $I\Sigma\_1+\Th\_{\Pi\_2}(\pa)$.


To see this, let $\phi$ be a $\Pi\_4$-sentence provable in $T$. Notice that $I\Sigma\_1+\Th\_{\Pi\_2(\pa)}$ is $\Pi\_3$-axiomatizable, hence there exists a $\Pi\_3$-sentence $\psi$ derivable in $I\Sigma\_1+\Th\_{\Pi\_2(\pa)}$ such that
$$\psi\land\neg\phi\vdash\rfn\_Q(\Pi\_4).$$
As before, an instance of $\rfn$ for the $\Pi\_4$-sentence $\neg(\psi\land\neg\phi)$ shows
$$\psi\land\neg\phi\vdash\mathrm{Con}\_{\psi\land\neg\phi},$$
hence $\psi\land\neg\phi$ is inconsistent by Gödel’s theorem, i.e.,
$$I\Sigma\_1+\Th\_{\Pi\_2}(\pa)\vdash\psi\vdash\phi.$$


[1] Kenneth McAloon, *On the complexity of models of arithmetic*, Journal of Symbolic Logic 47 (1982), no. 2, pp. 403–415.


[2] George Wilmers, *Bounded existential induction*, Journal of Symbolic Logic
50 (1985), no. 1, pp. 72–90.


