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32 votes
2 answers
3k views

Do we expect that sufficiently large computable ordinals settle every question of arithmetic?

I came across a post by Ron Maimon on physics.SE that makes what seems to me to be a very interesting conjecture I've never seen before about what it would take to settle every question of arithmetic. ...
Qiaochu Yuan's user avatar
19 votes
1 answer
748 views

What non-standard model of arithmetic does Hofstadter reference in GEB?

Following some of the coolest bits of Hofstadter's Gödel, Escher, Bach, extensions of the standard model of arithmetic are described. A ways in, the paragraph "Supernatural Addition and Multiplication"...
Dave Pritchard's user avatar
18 votes
3 answers
1k views

Computable nonstandard models for weak systems of arithmetic

By Tennenbaum's theorem, PA itself does not have any computable nonstandard models. The integer polynomials which are 0 or have a positive leading coefficient form a computable nonstandard model of ...
user avatar
18 votes
3 answers
2k views

Is Robinson Arithmetic biinterpretable with some theory in LST?

Let ZFC$^{\text{fin}}$ be ZFC minus the axiom of infinity plus the negation of the axiom of infinity. It is well-known that ZFC$^{\text{fin}}$ is biinterpretable with Peano Arithmetic. In this sense ...
Adam's user avatar
  • 3,267
17 votes
7 answers
2k views

Non-constructive proofs of decidability?

Are there examples of sets of natural numbers that are proven to be decidable but by non-constructive proofs only?
16 votes
3 answers
19k views

Non-computable but easily described arithmetical functions

I have read about the existence of functions of the kind described in the title in several places, but never seen an instance of them. Sorry if this is too much an elementary question to be posted ...
Marc Alcobé García's user avatar
16 votes
1 answer
831 views

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

By Tennenbaum's theorem, in the usual sense of computability for models, neither addition nor multiplication can be computable in a countable non-standard model of PA. Weak version: Can addition or ...
user avatar
12 votes
1 answer
835 views

Transfinitely extending $\sf PA$ — can we get stronger than $\sf ZFC$?

Let $\sf PA$ denote the theory of natural numbers with constants $(0, 1)$ and binary operators $(+,\times)$ based on the first-order predicate calculus with equality, having the following axioms, ...
Vladimir Reshetnikov's user avatar
12 votes
0 answers
249 views
+50

Is there a decidable theory of arithmetic with a non-collapsing quantifier hierarchy?

This question is very close to this old MSE question of mine, which is still unanswered. Is there an (ideally reasonably-natural!) expansion of the structure $(\mathbb{N};+)$ in a finite language ...
Noah Schweber's user avatar
11 votes
1 answer
400 views

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
10 votes
3 answers
618 views

If an oracle Turing machine halts with every infinite arithmetic oracle, can it fail to halt with some non-arithmetic oracle?

Let $e$ be an index of an oracle Turing machine program and $k$ be some natural number. Let us say that a subset of $\mathbb N$ is arithmetic if it is definable in the model $\langle \mathbb N,+,\cdot,...
Victoria Gitman's user avatar
7 votes
1 answer
198 views

A well-behaved $A$ that is almost contained in every element of some filter for a countable arithmetically closed family $\mathfrak X$

The question has relevance for constructing Scott sets with certain extra desirable properties. Suppose that $\mathfrak X$ is a countable arithmetically closed family of subsets of $\mathbb N$: ...
Victoria Gitman's user avatar
7 votes
1 answer
597 views

Can an uncountable model of Peano Arithmetic be recursive?

Can an uncountable model of Peano Arithmetic be recursive? What does it mean for an uncountable model to be recursive? Well, we represent the elements of the model using real numbers instead of ...
Christopher King's user avatar
7 votes
0 answers
110 views

How tightly are decidability and "induction-completeness" linked?

It is known that there are a number of expansions of the structure $\mathfrak{N}:=(\mathbb{N};+)$ which are decidable (= have computable theories); one such example is the expansion by a predicate ...
Noah Schweber's user avatar
6 votes
1 answer
197 views

Is there a 1-generic degree g such that Th(D(< g)) is more complicated than true arithmetic?

I am currently reading an article titled "Embedding and Coding Below a 1-Generic Degree" by Greenberg and Montalbán(link to a free source:https://pi.math.cornell.edu/~erlkonig/Papers/...
H.C Manu's user avatar
  • 893
5 votes
1 answer
195 views

Can extensions of $Q$ contradict Löb with recursive reflection?

It is an odd and arguably unacceptable situation that $PA$ does not have $\vdash_{PA}(Pr_{PA}\ulcorner A\urcorner\to A)$ for false recursive sentences $A$. However, it is not clear to me that Löb'...
Frode Alfson Bjørdal's user avatar
5 votes
1 answer
283 views

Which arithmetical sentences have no counterexamples in the sense of Kreisel?

It is a well-known fact that given a first-order sentence $\psi$ in prenex normal form $\forall x_1 \exists y_1 \forall x_2 \exists y_2 \dots \forall x_n \exists y_n \theta(x_1,\dots,x_n,y_1,\dots,y_n)...
James E Hanson's user avatar
5 votes
0 answers
109 views

Computational complexity of arithmetic sentences over classical theories

Below, I use the term "tracker" rather than "realizer" since I'm not requiring the relevant objects to be computable. Define the relation "$f$ tracks $\varphi$" for $f:\...
Noah Schweber's user avatar
5 votes
0 answers
318 views

$\Sigma_n$-complete sets in the Levy hierarchy

Recall that a set $A \subseteq \mathbb N$ is (many-one, Turing) $\Sigma_n$-complete if it's $\Sigma_n$ and any other $\Sigma_n$ set (many-one, Turing) reduces to it. This definition actually makes ...
Corey Bacal Switzer's user avatar
4 votes
1 answer
155 views

Does this hierarchy of fragments of $I \Sigma_1$ collapse?

Does anyone know whether the following hierarchy of fragments of $\mathrm{I} \Sigma_1$ (or rather $\mathrm{I} \Pi_1$) collapses or not? Let $\Sigma^b_n$ denote formulas in the language of arithmetic ...
Lukas Holter Melgaard's user avatar
4 votes
1 answer
258 views

What is the theory of statements with a provably *bounded* realizer (according to PA)?

$\let\T\mathrm\def\kr{\mathrel{\mathbf r}}$This is a follow up to Kleene realizability in Peano arithmetic. We can summarize the results from Emil Jeřábek's answer as follows: \begin{gather*} T_1 = \{ ...
Christopher King's user avatar
4 votes
1 answer
193 views

Further research on relevant realizability etc

I just read Dunn's 1979 paper Relevant Robinson's Arithmetic, and the end especially caught my interest. Following the surprising role of constant functions in collapsing "relevant Q with zero&...
Noah Schweber's user avatar
4 votes
1 answer
439 views

Alternative proof of Tennenbaum's theorem

The standard proof of Tennenbaums's theorem uses the existence of recursively enumerable inseparable sets and is presented e.g. in Kaye [1, 2], Smith [3]. In the following, $\mathcal{M}$ will always ...
Léreau's user avatar
  • 211
4 votes
1 answer
245 views

Induction and nonstandard halting times of standard machines

For a nonstandard model of enough arithmetic - say, $\mathcal{N}\models I\Sigma_1$ - we can define the set of halting times of standard machines relative to $\mathcal{N}$: $$SH(\mathcal{N})=\{n\in\...
Noah Schweber's user avatar
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
4 votes
0 answers
292 views

the strength of saying "each sentence of true arithmetic has a recursive proof"

Let $PA_{\omega}$ be just like $PA$ except that $PA_{\omega}$-proofs can use any number of applications of the recursive $\omega$-rule. The recursive $\omega$-rule allows the following: For each ...
Haidar's user avatar
  • 449
3 votes
3 answers
683 views

Turing Functional and $\Sigma_1^0$-formulas in models of fragments of PA

In models of PA with restricted induction power (for example, only $I\Sigma_n$ is present), the failure of higher induction scheme is characterised by the existence of definable cuts (like $\Sigma_2$ ...
Jing Zhang's user avatar
  • 3,038
3 votes
1 answer
437 views

Axiomatizations of complete theories

This question was motivated by this recent question by Ricky Demer. In his paper $\Pi^0_1$ classes and Boolean combinations of recursively enumerable sets, Carl Jockusch showed that there is no ...
François G. Dorais's user avatar
3 votes
1 answer
746 views

Turing degrees of nonstandard models of PA

Since the theorems of (PA + "there is a nonstandard number") are recursively enumerable, by the Low Basis Theorem, WKL0's proof of the completeness theorem gives a nonstandard model of PA of [low ...
user avatar
3 votes
1 answer
123 views

Kleene normal form theorem for r.e. relations proven in arithmetical theories

After reading the relevant chapters of Classical Recursion Theory (freely available from here), I have the following questions concerning Theorem II.1.10 (Normal form theorem) and Theorem IV.1.9 (...
CBuch's user avatar
  • 31
3 votes
0 answers
85 views

What is the relation of total functions in second order arithmetic and fast growing hierarchies?

Answer to this questions shows that fast growing hierarchies can grow arbitrarily fast for some definition of 'arbitrary'. Can second order arithmetic define all these functions (for any ordinal) ...
Łukasz Lew's user avatar
3 votes
0 answers
324 views

Is the quantifier-free fragment of Robinson arithmetic essentially undecidable?

It is well known that Robinson arithmetic (Q) is undecidable, and in fact essentially undecidable. Matiyasevich's theorem implies that the quantifier-free fragment of Q is also undecidable. However, I'...
Mak Nazečić-Andrlon's user avatar
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
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
  • 18.7k
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
1 vote
0 answers
123 views

Is possibile to define transfinite sum and product recursively? [closed]

On mathstackexchange a few days ago I published the following question where I asked about "transfinite" sum and products but actually nobody answered or gave an opinion with a comment: thus ...
Antonio Maria Di Mauro's user avatar
1 vote
0 answers
148 views

Why doesn't $\mathsf{B}\Sigma_2$ hold in $\mathsf{RCA}_0$?

For a formula $\varphi(i,u)$ of arithmetic, the bounding principle for $\varphi$ is the statement $$\forall m \, \Big( \big( \forall i<m\ \exists u\ \varphi(i,u) \big) \to \big( \exists v\ \forall ...
Jordan Barrett's user avatar
1 vote
0 answers
205 views

Lowest Turing degree that allows a Turing machine to tell whether $\operatorname{Con}(PA)$?

Let $T$ be a given turing machine. We say that $T$ decides $\operatorname{Con}(PA)$ if $PA + \operatorname{Con}(PA) \vdash T \text { accepts}$ and $PA + \lnot \operatorname{Con}(PA) \vdash T \text { ...
Christopher King's user avatar
0 votes
1 answer
336 views

Definability of arithmetic functions and relations

Motivation: Many "weak" arithmetic functions and/or relations ("relations" for short) are equivalent with relations explicitly definable by relations which were recursively defined by them beforehand (...
Hans-Peter Stricker's user avatar