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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
  • 35.5k
12 votes
0 answers
248 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
0 answers
476 views

Which sentences are "irreducibly" self-referential over $\mathsf{PA}$?

Previously asked at MSE. Below, all sentences/formulas are in the language of arithmetic, and for simplicity we conflate numbers with numerals and sentences with Godel numbers. Say that a sentence $\...
Noah Schweber's user avatar
10 votes
0 answers
161 views

Minkowski's lattice theorem in fragments of arithmetic

It is widely remarked that Minkowski's lattice theorem (or, convex body theorem) is a kind of geometrical pigeonhole principle. And it seems it should have a very elementary proof at least for convex ...
Colin McLarty's user avatar
9 votes
0 answers
210 views

Is there an Arithmetized Completeness theorem for intuitionistic theories?

For classical theories, Henkin's completeness proof can be arithmetized. This leads to the result that for classical theories $T$ and $S$ if $\sigma$ is a formula enumerating $S$ in $T$ then $S \leq T ...
Spencer Woolfson's user avatar
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
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
8 votes
0 answers
345 views

What arithmetic is interpretable in Mayberry's Euclidean set theory?

John Mayberry published what he calls a Euclidean set theory in his book The Foundations of Mathematics in the Theory of Sets. It is ZF with the axiom of infinity replaced by an axiom saying "the ...
Colin McLarty's user avatar
8 votes
0 answers
1k views

What's Reeb's take on naive integers?

Georges Reeb's "claim Q" is the statement that "naive integers don't fill up $\mathbb{N}$". To anyone familiar with model theory this could easily be interpreted as the existence of nonstandard models ...
Mikhail Katz's user avatar
  • 16.6k
8 votes
0 answers
198 views

Kripke models of $HA$

Let $K$ be a kripke model and $k$ be one of its node, then $\mathcal{M}_k$ is classical structure of $k$. What is the strongest theory of arithmetic like $T$ such that for every kripke model $K\...
Erfan Khaniki'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
7 votes
0 answers
284 views

Generic behavior of "polynomialish" models of $\mathsf{Q}$

(This question was originally asked and bountied at MSE - with different notation, some more explicit arguments, and topology in place of forcing.) Suppose $\mathcal{R}=(R_i)_{i\in\mathbb{N}}$ is a ...
Noah Schweber's user avatar
7 votes
0 answers
102 views

How similar can a model of $I\Delta_0$ be to the intersection of all of its definable cuts?

Let $M$ be a model of $I\Delta_0$. Recall that a definable cut is a definable (possibly with parameters) subset $I$ of $M$ that is non-empty, downwards closed, and closed under successor. If we ...
James E Hanson's user avatar
7 votes
0 answers
344 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
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
6 votes
0 answers
407 views

Can Set Theory be turned into Infinite Arithmetic?

The following system I'd label as "Infinite Arithmetic" is simply an endeavor to extend second order arithmetic to the infinite ordinal world, and extending with it the representation of ...
Zuhair Al-Johar's user avatar
6 votes
0 answers
192 views

How to show that $\omega^\omega$ is well-founded in PA?

By induction on $n$ variables I can show that for any meta-natural number $n$, PA proves well-foundedness of $\omega^n$. However it is well known that PA proves well-foundedness of $\omega^\omega$ ...
SmileLee's user avatar
  • 101
6 votes
0 answers
428 views

Proof of Tennenbaum's Theorem by McCarty

Tennenbaum's Theorem in its usual form states that for any countable non-standard model $M$ of PA there is no way to code the elements of $M$ as natural numbers such that either the addition or ...
Léreau's user avatar
  • 211
6 votes
0 answers
422 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
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
5 votes
0 answers
58 views

Mathematical strength of the statement "Heyting Arithmetic admits Markov's rule"

Consider the following theorem about Heyting arithmetic (HA) For every arithmetical formula $\phi$ whose only free variable is $n$, if $\text{HA} \vdash \forall n. \phi \lor \lnot \phi$ and $\text{HA}...
Christopher King'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
5 votes
0 answers
287 views

Is the two variable fragment of arithmetic, i.e., theory of ($\mathbb{N}, + ,\times$), decidable?

Any references would be appreciated. Most places only address different vocabularies (e.g. a survey of arithmetical definability by Bes).
Thinniyam Srinivasan Ramanatha's user avatar
4 votes
0 answers
162 views

Can this theory of dyadic rationals prove that multiplying by three is the same as summing thrice?

(This question arose from a discussion with Jade Vanadium about a theory of dyadic rationals.) Let $T$ be the 2-sorted FOL theory with sorts $ℕ,ℚ$ and constant-symbols $0,1$ and binary function-...
user21820's user avatar
  • 2,912
4 votes
0 answers
198 views

Is there a simple proof of consistency of EA?

Let $\mathsf{EA}+\mathsf{CE}$ be elementary arithmetic with cut elimination theorem. Is there a simple (1-)consistency proof of $\mathsf{EA}$ over $\mathsf{EA}+\mathsf{CE}$? I think that a naïve ...
Alwe's user avatar
  • 178
4 votes
0 answers
105 views

Computably saturated Skolem hulls of Morley sequences in $\mathsf{PA}$

Recall that a model $M$ of a first-order theory $T$ (in a computable language $\mathcal{L}$) is computably saturated if for every finite tuple $\bar{a} \in M$ and every computable partial type $\Sigma(...
James E Hanson's user avatar
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
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
203 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
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
4 votes
0 answers
219 views

Construction of model of arithmetic from an arbitrary model

Let $M$ be a non-standard model of $PA$, $a\in |M|$ be an arbitrary non-standard number and $T$ be a theory of arithmetic. We want to choose a subset $M'\subsetneq M$ such that: $M'\models PA^-$ (or $...
Erfan Khaniki's user avatar
4 votes
0 answers
126 views

Lascar strong types in fragments of arithmetic

Are Lascar strong types (definition below) in models of fragments of arithmetic always type definable? (They trivially are, in models of full induction.) Definition Given a saturated model ${\cal M}$ ...
Domenico Zambella's user avatar
3 votes
0 answers
206 views

Independence and truth in PA

By $\textbf{PA}$ I will mean the usual first-order Peano Arithmetic. I will denote an element of $\mathbb{N}$ by $n$, and by $[n]$ I will denote the corresponding term in the language of $\textbf{PA}$:...
jg1896's user avatar
  • 3,318
3 votes
0 answers
283 views

What is the meaning and proof of Harvey Friedman’s ultrafinite incompleteness sentence?

On page 7 of his paper “Adventures in Incompleteness”, Harvey Friedman states the following: IN ANY LONG ENOUGH SEQUENCE $x_1,...,x_n$ FROM $\{1,2,3\}$, SOME $(x_i,...,x_{2i})$ IS A SUBSEQUENCE OF ...
Keshav Srinivasan's user avatar
3 votes
0 answers
210 views

Self-referential Quinean proof of Löb's Theorem

Given its relevance for Open-source game theory, Dr. Andrew Critch asks the following about provability logic: We conjecture that Löb’s Theorem can be proven without the use of the modal fixed point $...
Martín S's user avatar
  • 421
3 votes
0 answers
165 views

Why is the proof of decidability of arithmetic (Theorem 2.1) in Hamkins & Lewis (2000) enough?

Recently, I was reading the paper "Infinite Time Turing Machines" by Hamkins & Lewis. And one of the first theorems (Theorem 2.1) is about decidability of arithmetic. The proof is quite ...
Jeremy's user avatar
  • 31
3 votes
0 answers
160 views

Is anything known about $\Delta_n$ bounding?

For a class $\Gamma \in \{ \Sigma_n, \Pi_n, \Delta_n \}$ in the arithmetical hierarchy, we can consider the induction, bounding, and least number principles for $\Gamma$: $\mathsf{I}\Gamma$ is $\big[ ...
Jordan Barrett's user avatar
3 votes
0 answers
146 views

Does Robinson arithmetic interpret a Kripke model of the double negation translation of $\mathsf{I}\Delta_0 + \mathrm{Exp}$?

It is a well-known fact that while while Robinson arithmetic can interpret surprisingly strong theories, it cannot interpret $\mathsf{I}\Delta_0 + \mathrm{Exp}$, i.e., Peano arithmetic with induction ...
James E Hanson'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
  • 675
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
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
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
3 votes
0 answers
144 views

A conservativity result of intuitionistic set theory over arithmetic

In their 1985 paper "Arithmetic Transfinite Induction and Recursive Well-Orderings", Friedman and Ščedrov prove that the theory $\mathbf{ZFI}$ is conservative over $\mathbf{HA}^*$ (see here, Theorem ...
namsap's user avatar
  • 355
3 votes
0 answers
198 views

What is the known weakest axiom system has Löb's derivability conditions?

We know that Peano Arithmetic satisfies Löb's derivability conditions, which is required in the proof of Gödel's 2nd incompleteness theorem. Is this the best result? If not, is there any known weaker ...
Ruizhi Yang's user avatar
3 votes
0 answers
343 views

Peano (Dedekind) categoricity

What is the smallest fragment of second order logic such that $Th(\mathbb{N})$ in that logic is categorical (only one model, namely natural numbers, up to isomorphism). For example, can we do this in ...
Thinniyam Srinivasan Ramanatha's user avatar
3 votes
0 answers
343 views

example just slightly better than the greedy construction

Roth's theorem provides an estimate for the largest size of a nonaveraging subset of $\lbrace 1,2, \ldots ,n \rbrace$ (a set of integers is nonaveraging if it does not contain any nontrivial three-...
Ewan Delanoy's user avatar
  • 3,595
3 votes
0 answers
771 views

Why isn't Montgomery modular exponentiation considered for use in quantum factoring?

It is well known that modular exponentiation (the main part of an RSA operation) is computationally expensive, and as far as I understand things the technique of Montgomery modular exponentiation is ...
Steve Huntsman's user avatar
2 votes
0 answers
100 views

Realizing arithmetic hierarchy in algebraic number theory

Is it possible to realize arithmetic hierarchy in algebraic number theory? For example, consider a $\Pi^0_4$ statement of the form $\forall x \exists y \forall z \exists w \phi(x,y,z,w)=0$ where $\phi$...
0x11111's user avatar
  • 593