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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
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
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
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
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
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
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