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Theories of arithmetic in first-order logic, such as Peano arithmetic, second-order arithmetic, Heyting arithmetic, and their subsystems and extensions. Models of Peano arithmetic.

15 votes

What is the canonical way to extend Peano's axioms to the set of all integers?

Here is a somewhat different way to think about it, although the result is equivalent to the theories in the other answers. Begin with the observation that the structures $\langle\mathbb{N},+,\cdot,0, …
Joel David Hamkins's user avatar
4 votes

Does strong provability imply syntactical provability?

Let me assume that you also include the axioms $\vdash A$ whenever $A$ is a logical axiom, for some fixed standard proof system using modus ponens as the only inference rule. And let me also assume th …
Joel David Hamkins's user avatar
6 votes
Accepted

Do these two provability theories over PA differ in consistency strength?

No, both of these theories are equiconsistent with PA. The reason is that given any model of PA, we may expand it to a model in the language with $\vdash$ by interpreting $\vdash\psi$ as true in the m …
Joel David Hamkins's user avatar
9 votes
Accepted

Can we use remote provability to prove the first incompleteness theorem sans $\omega$-consis...

It doesn't contradict PA to have such a descending sequence. For example, you could instead simply keep substracting one from $g^{*1}$, and this would be provably descending, but there is no contradic …
Joel David Hamkins's user avatar
12 votes
Accepted

Is there a useful measure of density of decidable sentences in PA?

Asymptotic density seems a very natural measure. The density of a set of sentences is the limit as $n\to\infty$ of the proportion of those sentences of length at most $n$ amongst all sentences of leng …
Joel David Hamkins's user avatar
8 votes

Can set theory be interpreted in infinite arithmetic?

Without considering your system of arithmetic too closely, let me mention that ZFC is interpretable in Peano arithmetic, if one augments PA with the assertion that ZFC is consistent. That is, ZFC is i …
Joel David Hamkins's user avatar
16 votes

Con(PA) via non-well-foundedness?

This is a completely standard perspective in work on models-of-PA, a view that informs dozens of arguments. That is simply the nature of nonstandard models, that things they think are well founded are …
Joel David Hamkins's user avatar
7 votes

Interpreting peano arithmetic without parameters

Let me give a simple example, which may help to clarify things. Consider the integer line $\langle\newcommand\Z{\mathbb{Z}}\Z,<\rangle$, and the natural number order $\langle\newcommand\N{\mathbb{N} …
Joel David Hamkins's user avatar
8 votes

Can there be computable non-standard models of PA in a weaker sense?

This is a great question! Let me give a meager partial answer, for the case where we are talking about nonstandard models of true arithmetic. Theorem. No nonstandard model of true arithmetic arises …
Joel David Hamkins's user avatar
7 votes
1 answer
571 views

Finding a PA cut in a nonstandard model of PA

For a certain project I am currently working on, I need to be able to find PA cuts in nonstandard models of PA, in desirable intervals. For example, I wonder if the following is true, where $\newcomma …
Joel David Hamkins's user avatar
14 votes
Accepted

Dedekind-Peano axioms, but numbers have at most one successor

Victoria Gitman and I recently looked at the first-order version of this theory, which we called fPA, to contrast it with the theory FPA, which Woodin has looked into, which is a possibly stronger ver …
Joel David Hamkins's user avatar
7 votes

The additive structure of clusters of nonstandard models of arithmetic

If you intend literally to recover the addition operation of the given model $M$, then the answer is negative. For every nonstandard model of arithmetic $M$ there is another model $M'$ having exactly …
Joel David Hamkins's user avatar
7 votes
Accepted

The additive structure of clusters of nonstandard models of arithmetic

The answer is yes. The additive cluster structure knows the additive structure of the original model up to isomorphism. Theorem. The additive structure of any countable nonstandard model $M$ of $\newc …
Joel David Hamkins's user avatar
2 votes

Is diamond consistent with 2nd order PA?

In the case of $\text{GBC}^-$, the answer is no, $\Diamond$ is not provable from the assumption that all reals are constructible. It is consistent with $\text{GBC}^-$ and even $\text{KM}^-$ that $\Dia …
Joel David Hamkins's user avatar
11 votes
Accepted

Can the "real" Peano Arithmetic be inconsistent?

It seems that the Feferman-style description of PA will exhibit your requirements. Specifically, consider the theory $P$ defined as follows. Begin to enumerate the usual PA axioms, but include the n …
Joel David Hamkins's user avatar

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