All Questions
8 questions
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. ...
- 118k
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 ...
user5810
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, ...
- 6,829
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:\...
- 20.7k
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 = \{ ...
- 6,403
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)....
- 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 ...
- 20.7k