Unsolvability Degree in Turing's Proof 1

I have read that there is some debate over the exact origin of the Halting argument, which begins with Kleene and Davis in the 1950s [Copeland 2004]. Motivated by this I want to clarify the Degree of the problem actually proved unsolvable by Turing [1936]. Of course the Halting problem is ${\Delta^0_2}$, but it is based on Post's (1947) model rather than Turing's (1936) model with its circular and circle-free machines.

My own quick analysis of this is that we would translate as follows:

M circle-free = $\{m| \phi(m) converges\}$ is infinite

This represents the fact that a Turing a-machine M will converge in infinitely many F- squares producing a computable real (say). $\phi$ is the corresponding (partial) recursive function computed by a-machine M. The D.N of M Turing called Satisfactory.

M circular = $\{m| \phi(m) converges\}$ is finite

This machine converges only finitely often. The difference between "circular" and the "terminating" of the Halting proof, is that a circular machine might not actually stop, since computation will continue (on the E-squares). The basic example of this is within Turing's proof itself wherein a candidate machine prints N-1 values then loops in its attempt to print any more: hence the term "circular".

So Turing's Proof 1 (which we could call the Satisfactoriness problem) is to show that no Turing a-machine D can decide D(M) = 1 if M is satisfactory; D(M)=0 if M is circular.

My calculation is that this satisfactoriness problem for D is at Level $\Delta^0_3$.

Has anyone else done this calculation?

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