When writing with a friend of mine today we came up with idea of extending ITTM concept of Hamkins and Kidder. First of all, I am familiar with one of Hamkins and Lewis results saying that every machine either halts or repeats before $\omega_1$, so allowing standard ITTMs to run beyond $\omega_1$ will give us nothing (because machine either already halted or won't halt ever).

The idea we had is the following: standard ITTM introduces specialized limit state the machine enters at every limit stage, that is, at every stage which is a multiple of $\omega$. To extend this, we add a second limit stage, such that machine enters it at stages which are multiples of $\omega_1$.

How strong is this extended system? The answer is: very strong. To show that I'll show how to decide $0^\blacktriangledown$, the set of all pairs $\langle e,r\rangle\in\mathbb{N}\times\mathbb{R}$ such that $e$th ITTM halts on input $r$: we just simulate this machine on that input! If the machine halts on this input, we will eventually notice this, and we can accept. Moreover, if the machine halts, it halts before $\omega_1$. So if we reach stage $\omega_1$ we know machine won't halt. But with our new limit stage we know when we reach $\omega_1$. Thus $0^\blacktriangledown$ is decidable.

From this fact, by a result of P.D.Welch, $\Sigma$ and ordinals well beyond it are writable. We can also iterate the above argument, so we prove decidability of $0^{\blacktriangledown^n}$.

Even better, we can iterate $\blacktriangledown$ operator transfinitely, up to ordinals of size of $\Sigma$. The argument to show this is actually quite similar to Hamkins-Lewis argument to show that tranfinitely iterated Turing jump is ITTM-computable.

When figuring this all out, I kept thinking about one thing - the relation between this model and ITTMs is very similar to relation between ITTMs and standard Turing machines. I also was thinking "there is no way this hasn't been considered before!".

The actual question of mine: has this system been considered by anyone before? The extension seems so simple I'd be surprised if no one did. It also is a very strong system, so my second question is: if this has already been considered, how strong is this system, actually? I suspect its complexity to be properly contained in $\Delta^1_2$, like with normal ITTMs.

(a little off-topic question, but I don't want to make separate MO question: I read somewhere that Koepke's ordinal machines with no ordinal parameters compute exactly $\Delta^1_2$ sets. Does anyone know good reference for this?)


1 Answer 1


I introduced a very similar model in my paper Cardinal-Recognizing Infinite Time Turing Machines, where I add a state which fires at cardinal times. It is easy to see that your model and mine are computationally equivalent, although the precise values of the ordinals $\gamma,\lambda,\Sigma$ etc. will of course differ.

As I show in the paper, these machines are equivalent to ordinary ITTMs with a $0^\blacktriangledown$ oracle. In particular, they are still very much contained within $\Delta^1_2$. This result also makes me doubt your claim that your machines can compute higher iterates of the strong jump. Perhaps you meant something else?

  • $\begingroup$ After re-thinking my argument, I have realized where it fails. Namely, we cannot clock things like $\omega_1^2$, so we can't know if the simulation of machine with $0^\blacktriangledown$ has already made enough steps so that it will never halt. $\endgroup$
    – Wojowu
    Commented May 29, 2014 at 13:59
  • $\begingroup$ In your paper, you put the question "does $\Sigma=\Sigma_C$?". This is false, because $\Sigma$ is computable from $0^\blacktriangledown$, as shown by Welch here (it's proven in proof of Corollary 3.6) $\endgroup$
    – Wojowu
    Commented May 29, 2014 at 14:02
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    $\begingroup$ @Wojowu Yes, the arXiv version is a bit out of date. At one point Anton Pertun showed me an argument, based on the one you pointed to, which solved this question, among other things. And even before that Bob Lubarsky gave a more complicated solution. $\endgroup$ Commented May 29, 2014 at 17:59

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