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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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 $\Sigma_2$ ...