Presburger arithmetic is consistent, but relative to what?

Question

In the Wikipedia article for (the first-order theory of) Presburger arithmetic, it is stated (among other properties) that Presburger arithmetic is consistent. What meta-theory does he rely on in order to prove this is consistent? Is that meta-theory itself "obviously" consistent? I realise I should read the paper, but I hope someone could give me a summary.

I've seen elsewhere people say things like: Presburger arithmetic is obviously consistent, because the natural numbers provide a model. Is this Presburger's argument? Does this argument work for Peano and Robinson arithmetic too? Does this line of reasoning not beg-the-question? To my mind the question of the consistency of the intuitive idea we have of the natural numbers (the thing we're using as a model) is one of the reasons we write things down formally in (for example) first-order logic.

Here is a bit of context of my thinking. Presburger arithmetic does not have the strength (being weaker than Peano arithmetic) to prove its own consistency - although that would not be very satisfying anyway, as we're calling into question the consistency of Presburger to begin with. If a theory is inconsistent, then it can prove anything. Including any statement asserting the theory's consistency - so long as you can write the sentence in the language of the theory. So, it seems to me that the best one can hope with any theory are relative consistency results. Since, the system you use to carry out a consistency proof must itself be proven consistent. Ad infinitum.

Thanks in advance :)

It seems to me that consistency is not discussed into Presburger's original paper and Presburger arithmetic does NOT prove its own consistency. If so, a meta-mathematical proof is needed, like that provided by Gentzen's for $\mathsf {PA}$. — Mauro ALLEGRANZA, Aug 24 '22 at 06:31
@MauroALLEGRANZA thanks for clarifying that. Like I said, I should have looked at the paper. Not just Wikipedia. — , Aug 24 '22 at 09:11
It's not so much that Presburger arithmetic does not have the strength to prove its own consistency as that it does not even have the ability to express its own consistency. — Eric Wofsey, Apr 29 '23 at 00:45

score 3 · Accepted Answer · answered Apr 29 '23 at 00:54

You are right that there is no completely philosophically satisfying way to prove a foundational theory's consistency--however you prove it, that will just raise the question of the soundness of the metatheory you used to prove it. You can't get something from nothing--at some point you just have to accept some base theory to use as a foundation to prove everything else. When people talk about a theory being consistent, they generally just mean its consistency can be proved from some generally accepted foundation for mathematics. For instance, the existence of the natural numbers as a model immediately implies the consistency of Presburger arithmetic (or of stronger theories of arithmetic like PA) can be proved in ZF (or even a much weaker set theory). Probably it can even be proved in just PA (apparently Presburger gave an algorithm to decide whether Presburger arithmetic can prove a given sentence, and probably his proof of the correctness of this algorithm can be carried out in PA), but I don't know the details to say for sure.

PA is massive overkill for the consistency of Presburger (for each $n$ PA proves $Con(I\Sigma_n)$, after all). — Noah Schweber, Apr 29 '23 at 05:31

Presburger arithmetic is consistent, but relative to what?

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