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?
 A: I am answering this question that I posted in 2012 since there were no responses. The short answer is "Yes" - the calculation has been done before; also some papers published in the 2012/2013 period (discussed below) add some further results to this topic.
In his article in the publication (The Universal Turing Machine, R.Herken, OUP 1988) celebrating the 50th anniversary of the Turing Machine, Davies is discussing the 1936 paper argument. In a footnote 5 he makes the following comment:


  
*A remark for the knowledgeable: the problem of determining whether a given Turing machine is circle-free is complete of degree $\mathbf{0''}$ ...
  

Now the a theorem of Recursion theory gives:
$\Delta_{n+1} = \mathbf0^{(n)}$
So the circle-free condition is $\Delta_{3}$ as indicated in the question.
The significance of this result is that the Diagonalisation lemma is stating what is excluded for a Turing Machine: it cannot solve $\Delta_{3}$ problems.
Of course the Halting theorem is based on the alternative Post (1936, 1947) model of computation and Halting is $\mathbf{0'}$ complete - one Turing degree lower than Turing's own result demonstrated. Thus a Post machine cannot solve 
$\Delta_{2}$ problems - the familiar kind of "non-algorithmic" functions discussed by Penrose and textbooks generally.
Thus Turing's own result is actually showing that his machines are more powerful than the "Textbook Turing Machine". This idea has been exploited indirectly and latterly more directly in papers on extended computation. An example is the paper published in 2013:
"The Computational Power of Turing's Non-terminating Circular A-Machines" 
 (here ) by van Leeuwen and Wiedermann in Alan Turing His Work and Impact.
The first sentence in this paper begins:

For readers familiar with the concept of Turing machines as described in contemporary textbooks, reading the definition of a Turing machine in Turing's original paper (Turing 1936) may present a surprise.

After I found this result I was myself surprised that another icon of 1930s Logic was not all that it seemed as evidenced in my Stack Question from 2014 (Constructiveness of Proof of Gödel's Completeness Theorem).
