7

Let $\Sigma=\{+,<,0,1\}$ be the usual language of Presburger arithmetic - so we have addition but no multiplication (the additional symbols for $<,0,1$ are unnecessary but convenient). Given a "reasonable" logic $\mathcal{L}$, let $\mathbb{Pres}(\mathcal{L})$ be the $\mathcal{L}$-theory consisting of

  • Axioms 1-4 of the usual presentation of Presburger arithmetic, and

  • for each $\mathcal{L}[\Sigma]$-formula $\varphi(x,\overline{y})$, the $\mathcal{L}[\Sigma]$-sentence $$\forall \overline{y}[\varphi(0,\overline{y})\wedge\forall x(\varphi(x,\overline{y})\rightarrow\varphi(x+1,\overline{y}))\implies \forall x\varphi(x,\overline{y})].$$

For example, $\mathbb{Pres}(\mathsf{SOL})$ is fully categorical. $\mathbb{Pres}(\mathsf{FOL})$ (= usual Presburger arithmetic) is not fully categorical of course, but it is complete as an $\mathsf{FOL}$-theory: for each first-order sentence $\theta$ in the language $\Sigma$, either $\mathbb{Pres}(\mathsf{FOL})\models\theta$ or $\mathbb{Pres}(\mathsf{FOL})\models\neg\theta$.

I'm curious about whether there is a "natural" logic $\mathcal{L}$ such that $\mathbb{Pres}(\mathcal{L})$ is incomplete as an $\mathcal{L}$-theory. There are a couple natural candidates coming from a similarly-flavored analogue of this question for PA, and I think the simplest is "Ramsey logic," that is, first-order logic equipped with the Ramsey/Magidor-Malitz quantifiers (see e.g. this answer of Enayat):

Let $\mathcal{R}$ be first-order logic augmented by each Ramsey quantifier $Q^n$. Is $\mathbb{Pres}(\mathcal{R})$ complete as an $\mathcal{R}$-theory? That is, is it the case that, for every $\mathcal{R}[\Sigma]$-sentence $\theta$, either $\mathbb{Pres}(\mathcal{R})\models\theta$ or $\mathbb{Pres}(\mathcal{R})\models\neg\theta$?

Noah Schweber
  • 260,658
  • I am not sure this is correct, treat with caution. A logic that makes Pres$(\mathcal L)$ incomplete could be Fixpoint logic. In this logic you should be able to define multiplication and a sufficiently large fragment of PA. – Primo Petri Mar 20 '22 at 13:31
  • @PrimoPetri Can't fixpoint logic pin down $(\mathbb{N};+,\times)$ though? (So I'm worried you'd still get something complete.) – Noah Schweber Mar 20 '22 at 18:18
  • I did not check the details. Here is the heuristic: fixpoint logic capture PSPACE. This is sufficient to define multiplication and more. Together with your induction axiom, you should get $I\Delta_0+\Omega_0$ and more. But as PSPACE$\ \subseteq\ $EXP, the strength of the theory is limited by $I\Delta_0+$exp$\ \subseteq I\Sigma_1\subseteq PA$, Therefore it would be far from complete. (I am going back and forth between conplexity classes and subsystems of PA --- this may not be completely unproblematic.) – Primo Petri Mar 20 '22 at 19:40
  • @PrimoPetri I think your application of complexity results is incorrect. When we say "$\mathcal{L}$ captures $\mathsf{C}$" we mean (usually) that the classes of finite structures which are $\mathsf{C}$-decidable are exactly those which are axiomatizable by an $\mathcal{L}$-sentence. That's not relevant here. And unless I'm misunderstanding, fixed-point logic can define the set of standard elements of a nonstandard model of arithmetic. – Noah Schweber Mar 20 '22 at 20:09
  • I'll give a thought about what you say. Just to check we are on the same page: the correct way to translate fixed-point logic into arithmetic is to apply it to bounded sets. Otherwise you get something way too strong and loose all connections with complexity classes. – Primo Petri Mar 21 '22 at 06:55
  • @PrimoPetri OK, what definition exactly are you using of "fixed-point logic" here? – Noah Schweber Apr 11 '22 at 17:41
  • Indeed, the connection between (pseudo)finite models and (bounded)arithmetic is worth spelling out. (It is going to take a few days, though.) – Primo Petri Apr 12 '22 at 12:41

0 Answers0