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In this Paper of D. Isaacson, it is proved that the true arithmetic($Th(\mathbb{N}$)) is the only $\omega$-consistent and complete extension of $Q$ (Robinson's arithmetic). This, together with the fact that $Th(\mathbb{N})$ is not definable, immediately implies the following result:

Proposition. If $T\supseteq Q$ is a $\omega$-consistent theory such that the set of (Godel numbers of ) axioms of $T$ is definable by a $\Sigma_n$ (or $\Pi_n$) formula (for some $n\in \mathbb{N}$), then $T$ is incomplete.

It can be proved that this incompleteness phenomenon is essentially non-constructive, which means that there is no computable function $f$ such that for every formula $\sigma(x)$ which defines the set of (Godel numbers of) axioms of an $\omega$-consistent theory $T$, then $f(\ulcorner\sigma(x)\urcorner)\downarrow=\theta$ and $\theta$ is a sentece independent from $T$. It is the case even when we restrict the problem to $\Sigma_4$ formulas. The idea of the proof is as follows:

$\omega$-consistency of a r.e. theory can be written as a $\Pi_3$ formula. By using the parametric version of the diagonal lemma, we can construct a formula $\psi(x)$ such that : $Q\vdash \psi(x)\equiv [f(\ulcorner \psi \urcorner)\downarrow \wedge \omega\text{-}con(Q+f(\ulcorner \psi \urcorner)) \wedge (x=q \vee x=f(\ulcorner \psi \urcorner))] \vee$ $[f(\ulcorner \psi \urcorner)\downarrow \wedge \omega\text{-}con(Q+\neg f(\ulcorner \psi \urcorner)) \wedge (x=q \vee x=\neg f(\ulcorner \psi \urcorner))]\vee$ $[x=q]$

Where $q$ is the Godel number of conjunction of all axioms of $Q$. Now it is not hard to check that $\psi(x)$ is a $\Pi_3$ and then $\Sigma_4$ formula which defines a $\omega$-consistent theory $T_\psi$, but $f(\psi)$ is not independent from $T_\psi$ (see this preprint for more details).

So there is no computable $f$ with the desired property, even when we restrict the problem to $\Sigma_4$-definable theories. My question is about Turing degree of a function like $f$. Obviously one can enumerate the graph $f$ by oracle $0^{(4)}$ (because for a given $\Sigma_4$-definable theory, the problem of provability or refutability of a given sentence can be decided by oracle $0^{(4)}$). Hence it is a $0^{(4)}$-r.e. set, so its Turing degree is less that (or equal to) $0^{(5)}$.

Question: Is this Turing degree (necessarily) $0^{(5)}$ or it can be strictly less than $0^{(5)}$?

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  • $\begingroup$ Interesting question! $\endgroup$ – Shahrooz Janbaz May 22 '16 at 7:51
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From your use of "$\downarrow$" it seems you do not require $f$ to give an answer on non-$\omega$-consistent theories.

At least we can get a function $f$ partial computable in $0^{(3)}$ (perhaps you already knew this).

We just go through the $\Sigma_3$ formulas $\sigma$ defining the axioms of theories $T$, and ask, given an input $\theta$,

Is there a proof or disproof in $T$ of $\theta$?

This is a $\Sigma^0_3$ property so $0^{(3)}$ can give a definite answer yes/no. If no, we let $f(\sigma)\downarrow=\theta$. If yes, we move on to the next $\theta$.

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  • $\begingroup$ Yes, you are right....but what I had in mind was Turing degree of the function $f$ as a r.e. set (i.e. Turing degree of its graph). Graph of the function you described is $0^{(3)}$-r.e., but its Turing degree may be bigger than $0^{(3)}$. I edited the question to make this point clear. $\endgroup$ – Payam Seraji Jun 23 '16 at 21:07
  • $\begingroup$ Sorry, there was a mistake and some 3's should be in fact be 4 and the last 4 should be 5....I corrected them. $\endgroup$ – Payam Seraji Jun 23 '16 at 21:44

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