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Questions tagged [theories-of-arithmetic]

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.

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What is the Turing degree of the monadic theory of the real line?

The monadic theory of the real line is the set of all sentences in the monadic second-order language of order which are true in $\mathbb{R}$. In this 1982 paper, Gurevich and Shelah show that true ...
Keshav Srinivasan's user avatar
16 votes
2 answers
1k views

How special is first-order $\mathsf{PA}$?

This is a modified version of a question which was asked and bountied at MSE without success. Below, "$\mathsf{PA}$" refers to first-order Peano arithmetic. There are various "...
Noah Schweber's user avatar
5 votes
1 answer
270 views

Can a nonstandard model of $\mathsf{PA}$ be "$\Delta^1_1$-well-ordered?"

This was asked and bountied at MSE with no response: My question is the following: Is there a nonstandard model $\mathcal{M}\models\mathsf{PA}$ such that $\mathcal{M}$ has no $\Delta^1_1$-with-...
Noah Schweber's user avatar
8 votes
1 answer
282 views

Paris-Harrington principles parametrized by functions $f:\mathbb N \to \mathbb N$

Recall that the Paris-Harrington Principle, $\mathsf{PH}$, is the statement that for each $e, r, k < \omega$ there is an $N < \omega$ so that given any coloring $c:[N]^e \to r$ there is an $H \...
Corey Bacal Switzer's user avatar
9 votes
1 answer
493 views

What can $I\Delta_0$ prove?

What combinatorial and number-theoretic propositions can $I\Delta_0$ prove? Obviously there are an infinitude of them, but what are some well known theorems that can be proved in $I\Delta_0$, if any?
BPP's user avatar
  • 675
9 votes
0 answers
204 views

Reverse mathematics of Noetherian rings over $\mathbb{Q}$

Take the Hilbert Basis Theorem over the rational numbers in this form in the language of Second Order Arithmetic:  For every $n\in N$ every ideal of the polynomial ring $\mathbb{Q}[x_1,\dots,x_n]$ is ...
Colin McLarty's user avatar
17 votes
3 answers
3k views

Did Edward Nelson accept the incompleteness theorems?

Edward Nelson advocated weak versions of arithmetic (called predicative arithmetic) that couldn't prove the totality of exponentiation. Since his theory extends Robinson arithmetic, the incompleteness ...
BPP's user avatar
  • 675
4 votes
0 answers
553 views

Can Robinson arithmetic prove any interesting theorems?

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 ...
BPP's user avatar
  • 675
10 votes
1 answer
630 views

Is $\mathsf{R}$ axiomatizable by finitely many schemes?

Recall that $\mathsf{R}$ is the theory of arithmetic consisting of the quantifier-free theory of $(\mathbb{N};+,\times,0,1,<)$ together with, for each $k\in\mathbb{N}$, the sentence $$\forall x[(\...
Noah Schweber's user avatar
3 votes
0 answers
191 views

Set theories that are complete modulo finite-order arithmetic

In a previous question, I asked whether there can be effectively axiomatizable set theories (at least as strong as, say, ZF) that are complete modulo first-order arithmetic, to which the answer is no; ...
BPP's user avatar
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4 votes
0 answers
431 views

How can I prove that primitive recursion “preserves” representability in Peano Arithmetic?

I'm working on my thesis about Gödel's Incompleteness Theorems, and at some point I need to prove that the $\textsf{PA}$ system is able to represent all the recursive functions. By recursive function ...
Ranopano's user avatar
3 votes
0 answers
172 views

Interpretability of primitive recursive functions in Peano Arithmetic

Let $R$ be a set of defining equations for primitive recursive functions successively built up from $s, +, \cdot$. Is PA + $R$ interpretable in PA? (Interpretability understood in the sense of Tarski, ...
Ansten Klev's user avatar
7 votes
0 answers
343 views

Nelson's contradiction in finitism

I have read up, in Shoenfield and elsewhere, on a lot of the details involved in Nelson's failed proof of the inconsistency of arithmetic. I understand the Kritchman-Raz proof; the proof of the ...
Jori's user avatar
  • 189
1 vote
0 answers
107 views

Formalization in PA in the Kritchman-Raz proof

In their paper Kritchman and Raz present a proof of Gödel's second theorem using Kolmogorov complexity. To make it work, they operate in some (weak) formal theory $T$ that incorporates some arithmetic,...
Jori's user avatar
  • 189
-1 votes
2 answers
638 views

Peano axioms— mathematical induction and other axioms

The Peano axioms of $\Bbb N$ are: $1 \in \Bbb N$, i.e. $\Bbb N$ is not empty and contains an element denoted by $1$. Every natural number has a successor, i.e. $\forall n\in\Bbb N, \exists!s(n)\in\...
Curiosity's user avatar
  • 101
2 votes
0 answers
237 views

Representing iteration of a function in PA

Let $\mathscr{L}$ be a (recursive) FOL language, with numeral symbols $\underline{0},\underline{1},\ldots$. Let $T$ be a recursive, consistent theory, containing PA (or even just Robinson arithmetic)....
Pace Nielsen's user avatar
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8 votes
1 answer
574 views

Iterated Gentzen: or, a Sith objection to the proof of consistency of PA

$\DeclareMathOperator\PRA{PRA}\DeclareMathOperator\WF{WF}\DeclareMathOperator\Con{Con}\DeclareMathOperator\PA{PA}$Preamble: In the year … in a galaxy far far away, a nasty Sith named Darth Dubious (...
Mirco A. Mannucci's user avatar
10 votes
1 answer
350 views

An internalized version of Tennenbaum's Theorem

Tennenbaum's celebrated 1959 theorem (see here for a reference) is certainly one of the key theorems in mathematical logic. Not so much for its proof, but because it helps "isolating" $N$ ...
Mirco A. Mannucci's user avatar
1 vote
1 answer
395 views

Complete and consistent first-order theories that contain interesting phenomena

Gödel has shown that a consistent recursively axiomatizable first-order theory that can interpret Robinson arithmetic is incomplete. I think there is some sentimental value in working with a theory ...
user avatar
10 votes
1 answer
414 views

Concrete examples of statements not provable in PRA + $\epsilon_0$-induction that are provable in PA?

It is well-known that $\mathbf{PRA}$ plus $\epsilon_0$-induction on bounded formulas cannot prove all $\mathbf{PA}$ theorems (essentially because $I\Sigma_1$ plus $\epsilon_0$-induction on bounded ...
Jori's user avatar
  • 189
5 votes
3 answers
1k views

Are there first-order statements that second order PA proves that first order PA does not?

Are there first-order statements that second order PA proves that first order PA does not? Is this known one way or the other? Could you share an example? (edit: to clarify, by 'second order PA' I don'...
pathway's user avatar
  • 117
4 votes
0 answers
202 views

The Return of Graham Arithmetics: adding induction up to $g_{64}$

In my previous question The inconsistency of Graham Arithmetics plus $ \forall n, n < g_{64}$, I introduced an extension of Robinson Arithmetics with the recursive definition of Tetraction, a small ...
Mirco A. Mannucci's user avatar
9 votes
0 answers
325 views

Is there a definable model of PA whose domain is a proper class and whose complete theory is not definable?

Assume ZFC. Is there a formula of $\mathcal{L}_\in$ (without parameters) defining a model $\mathcal{M}$ of PA whose domain is a proper class but the complete theory of that model is not definable by ...
Guy Crouchback's user avatar
12 votes
2 answers
866 views

The inconsistency of Graham Arithmetics plus $ \forall n, n < g_{64}$

As you all know, Ronald Graham just passed away. He is famous for many fabulous contributions to finite combinatorics, and much much more, but perhaps none of them is as popular as the infamous ...
Mirco A. Mannucci's user avatar
7 votes
0 answers
179 views

The provability logic of $I\Delta_0+\Omega_1 $ and complexity theory

Almost 30 years ago, a number of folks in provability logic tried to show that GL (see for instance the excellent survey by Rineke Verbrugge here) is indeed the logic of $I\Delta_0+\Omega_1$ (in the ...
Mirco A. Mannucci's user avatar
10 votes
1 answer
806 views

Why can't we embed Tarski's truth in PA?

I recently learned that ZFC can prove $Con(PA)$ because it can give a model of PA, but I'm not given the technical details. (My teacher thinks it is too obvious to even mention.) What plagues me is ...
Paul Sohn's user avatar
  • 171
5 votes
1 answer
385 views

Lob theorem for Robinson arithmetic

If i'm not wrong, the theory which Lob theorem applies to should be sufficiently strong, satisfying 3 "derivability" conditions, like PA. $Q$ is the Robinson arithmetic. I'm afraid $Q$, is ...
Ali's user avatar
  • 103
2 votes
2 answers
436 views

Is there any reasonable non-regular Gödel numbering of the language of arithmetic?

Let $\mathcal{L}$ be the language of arithmetic given as follows: $x::= {\sf v} \mid x'$ $t ::= x \mid 0 \mid {\sf S}t \mid (t+t) \mid (t\times t)$ $A ::= \bot \mid \top \mid t=t \mid \neg A \mid (A \...
Balthasar Grabmayr's user avatar
54 votes
1 answer
3k views

In the two-person Killing the Hydra game, what is the winning strategy?

My question is which player has a winning strategy in the two-player version of the Killing the Hydra game? In their amazing paper, Kirby, Laurie; Paris, Jeff, Accessible independence results for ...
Joel David Hamkins's user avatar
8 votes
1 answer
491 views

Natural $\Pi_1$ sentence independent of PA

Order invariant graphs and finite incompleteness by Harvey Friedman gives an example of a combinatorial/non-metamathematical $\Pi_1$ sentence that is independent of ZFC. Is there a simpler example of ...
user76284's user avatar
  • 2,203
2 votes
1 answer
273 views

Definability in countable nonstandard models of Peano arithmetic

I know that every element of $\mathbb{N}$ is definable the standard model of Peano Arithmetic. Does there exist a countable non-standard model of PA where the same is true?
Marcus Dubious's user avatar
16 votes
2 answers
2k views

Could Kronecker accept a proof of Goodstein's theorem?

A famous result of Goodstein asserts that the Goodstein sequence of integers terminates. For a precise statement and a short proof, see https://en.wikipedia.org/wiki/Goodstein%27s_theorem. A well ...
Piotr Hajlasz's user avatar
6 votes
2 answers
435 views

Interpreting proper elementarily equivalent end extensions?

Is there a tuple of parameter-free formulas $\Phi$ and a nonstandard $M\models PA$ such that $\Phi^M\models PA$, the induced $M$-definable initial segment embedding $j_\Phi^M:M\rightarrow\Phi^M$ is ...
Noah Schweber's user avatar
1 vote
2 answers
267 views

The "higher topology" of countable Scott sets

Fix some computable bijection $b$ between $\omega$ and $2^{<\omega}$. For $r\in 2^\omega$, let $$[r]=\{f\in 2^\omega: \forall\sigma\prec f(b^{-1}(\sigma)\in r)\}$$ be the closed subset of Cantor ...
Noah Schweber's user avatar
3 votes
0 answers
301 views

What does second order set theory give us that is new?

There is a natural analogy between the theories PA and ZFC. See the linked question by Gro-Tsen here. Peano arithmetic (PA) is a first order approximation to the natural numbers. As is well known, ...
Pace Nielsen's user avatar
  • 18.7k
9 votes
1 answer
643 views

Gentzen's result on PA

The Wikipedia states that Gentzen proved that "in modern terms, the proof-theoretic ordinal of PA is $\varepsilon_0$." Further down, the article defines what the "proof theoretic ordinal" of a theory ...
Gabriel Nivasch's user avatar
6 votes
0 answers
420 views

What is proof-theoretic ordinal of weak first-order arithmetic?

According to Wikipedia(https://en.wikipedia.org/wiki/Ordinal_analysis) and nlab(https://ncatlab.org/nlab/show/ordinal+analysis), a proof-theoretic ordinal of $\mathsf{PRA}$ is $\omega^\omega$. ...
Alwe's user avatar
  • 178
3 votes
1 answer
171 views

Models of arithmetical theory R + induction in which successor is not injective

Consider the arithmetical theory sometimes denoted by $\mathsf{R}$. The non-logical vocabulary of $\mathsf{R}$ consists of '$0$', '$S$', '$+$' and '$\times$'. The axioms of this theory are all true ...
Thomas Schindler's user avatar
1 vote
1 answer
269 views

Interpreting PA2 in second-order logic + existence of infinitely many objects

I've heard that if you assume the existence of (Dedekind) infinitely many objects, you can derive -- in second-order logic, given suitable definitions -- the (second-order) Peano axioms for arithmetic....
Thomas Schindler's user avatar
0 votes
3 answers
1k views

Regarding Gentzen's note regarding 'Godel-points' (i.e., "Where is the Godel-point hiding?")

Consider the following note written by Gerhard Gentzen in early 1932, on the onset of his work on a consistency proof for arithmetic: The axioms of arithmetic are obviously correct, and the ...
Thomas Benjamin's user avatar
2 votes
1 answer
365 views

Is second-order logic *with standard semantics* necessary to categorically characterise the natural number structure?

Is second-order logic with standard semantics necessary to categorically characterise the natural number structure? One can prove that any two models of Dedekind-Peano arithmetic are isomorphic (...
sdlkgjh45's user avatar
1 vote
0 answers
193 views

Induction on open formulas vs. Induction on $\Pi_1$ formulas

There are infinitely many extension to Robinson's $Q$ arithmetic many of which are defined by adding an axiom schema of induction for particular set of formulas. I am confused about the theory $\text{...
Punga's user avatar
  • 173
1 vote
1 answer
466 views

What does "can almost be proven in PA" mean regarding Theorem 2 of Timothy Chow's expository article, "The Consistency of Arithmetic"?

In his expository article, "The Consistency of Arithmetic" (MSN), Prof. Chow has the following theorems: Theorem 1. If $a_1, a_2, a_3,\dotsc$ is a sequence of ordinals and $a_i \ge a_j$ whenever $...
Thomas Benjamin's user avatar
6 votes
0 answers
113 views

When can two elementary end extensions of models of PA be uniquely amalgamated?

$\DeclareMathOperator{Cod}{Cod}$ $\DeclareMathOperator{Scl}{Scl}$ $\DeclareMathOperator{Def}{Def}$ $\DeclareMathOperator{Lt}{Lt}$ Background: All of the background to this question can be found in ...
Athar Abdul-Quader's user avatar
3 votes
1 answer
233 views

Is cyclic PA interpretable in PA?

If we remove the axiom that zero doesn't have a predecessor, and stipulate that every natural number has a predecessor, and that no number is the successor of itself. And keep all other axioms of $\...
Zuhair Al-Johar's user avatar
0 votes
0 answers
104 views

Multivariate polynomial with infinite but discrete roots on one variable

I want to know if there exists a polynomial $ P(z, x_1,x_2,\ldots,x_n)$ over the rationals such that the set $$ Z_P = \{z | \exists x_1,\ldots,x_n. P(z, x_1,x_2,\ldots,x_n) = 0 \} \subsetneq \mathbb Q ...
afiori's user avatar
  • 163
29 votes
10 answers
4k views

Defining the standard model of PA so that a space alien could understand

First, some context. In one of the comments to an answer to the recent question Why not adopt the constructibility axiom V=L? I was directed to some papers of Nik Weaver at this link, on ...
Pace Nielsen's user avatar
  • 18.7k
18 votes
1 answer
3k views

Existence of a model of ZFC in which the natural numbers are really the natural numbers

I know that, from compactness theorem, one can prove that there are models of first order arithmetic in which there is some "number" which is not a successor of zero, in the sense that it is strictly ...
display llvll's user avatar
7 votes
2 answers
241 views

Measure of the numbers with length of $n$ for a nonstandard number $n$

Is there any nonstandard model of $PA$ with the following properties? There exists a nonstandard number $n\in M$ such that $M\upharpoonright n$ is countable, Let $|x|=\lceil\log_2x\rceil$, then $|\{...
Erfan Khaniki's user avatar
14 votes
0 answers
654 views

Reverse Mathematics of Euclid's theorem

Euclid's theorem that there are infinitely many prime numbers has multiple proofs, ranging from Euclid's original theorem that constructs a new prime from a finite list of such, to Euler's proof that ...
David Roberts's user avatar
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