4
$\begingroup$

The motivation for my question is I'm curious whether studying Robinson arithmetic can be fruitful in the same sense as studying group theory. Robinson arithmetic is so weak that there are many structures that satisfy its axioms, just like there are many structures that satisfy the group axioms. But are there any interesting theorems we can prove about all such structures?

By interesting, I mean something that wouldn't be obvious to someone who has just learned how to count, but not because the statement is absurdly long and complex.

Since Robinson arithmetic is $\Sigma_1$-complete, I'm excluding $\Sigma_1$ statements.

Improve this question
$\endgroup$
7
  • $\begingroup$ Other than it's original usage and derivatives? Which version of Robinson arithmetic are you thinking of? $\endgroup$
    – François G. Dorais
    Commented Dec 8, 2020 at 22:38
  • 2
    $\begingroup$ As far as I understand, its original usage was for proving theorems about it, not for proving theorems within it. I'm thinking of the first-order theory whose axioms are those of PA without induction with the axiom that every number is either 0 or the successor of some number. $\endgroup$
    – BPP
    Commented Dec 8, 2020 at 23:00
  • $\begingroup$ "Those of PA" is the ambiguous part. If you're more specific you will see that there are lots and lots of models! (Especially if you stick to the basics and don't fall into the semiring trap.) I think they're interesting but I can only remember a couple of uses, both would perhaps seem sketchy to you. $\endgroup$
    – François G. Dorais
    Commented Dec 8, 2020 at 23:12
  • 3
    $\begingroup$ @DavidRoberts I think he is referring to the following. If I'm defining $\mathsf{PA}$ to an audience not interested in Robinson arithmetic, I'll say it's the nonnegative discrete ordered semiring axioms + the induction scheme. Throwing away induction then amounts to axioms saying that you're the nonnegative part of a discrete ordered ring. However, this is much stronger than Robinson arithmetic, which e.g. doesn't even prove that addition is commutative. The issue is that there are lots of ways to write $\mathsf{PA}=T+IndScheme$, which the slogan "$\mathsf{PA}-IndScheme=\mathsf{Q}$" ignores. $\endgroup$
    – Noah Schweber
    Commented Dec 9, 2020 at 0:12
  • 1
    $\begingroup$ I don’t know what counts as interesting, but Robinson’s arithmetic proves all kinds of things of the form “Bertrand’s postulate and Sylvester’s theorem hold for all numbers from this and this definable cut”. (The definition of the cut may be fairly long, sure.) $\endgroup$
    – Emil Jeřábek
    Commented Dec 9, 2020 at 8:12

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .