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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
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
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?
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
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
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
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
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