1 The Penrose-halting theorem

What we’ll call the ‘Penrose-halting theorem’ concerns the ‘self-halting’ problem: to compute whether any given computer will halt when its own machine number is input. To state and prove the theorem we first make some definitions.Footnote ¹

By Turing Machine (TM) I mean any mechanical computational device implementing an algorithm, within the scope of the Church–Turing thesis, not only those of the type explicitly introduced by Turing. Let T $_m$ be the mth TM in a particular computable enumeration of some class of TMs (assumed computationally equivalent to Turing’s).

For clarity of presentation I will use a ‘black box’ scheme for defining TMs: for instance, machines to compute $x+1$ and $2\times x$ could be represented by

respectively. Such boxes can be compounded to define new TMs, each component taking as its input the output of the preceding box. For instance, the TM Q defined as

obviously computes $2x+1$ .

Let $A[m$ ] denote the result of TM A acting on input $m\in \mathbb {N}$ : $A[m$ ] will either be the natural number output, or that $A[m]$ fails to halt (in which case the value of $A[m$ ] is undefined). So, for our example TM, $Q[m]=2m+1$ (defined for all $m\in \mathbb {N}$ ).

A partial self-halting computer (PSHC) is any TM with the following description:

where 0 and 1 are the only possible outputs. That is, if P halts then it accurately informs us whether or not T $_m$ is a ‘self-halter’. Of course, if P were to halt for all values of m then it would, per impossibile, solve the self-halting problem, so it must be the case that every PSHC fails to halt for some $m\in \mathbb {N}$ .Footnote ²

For any TM

define

where a diamond represents a decision based on its input (here, whether or not the input is 0), and loop is some specific routine that loops endlessly. Note that the only output that A* can have is 0; otherwise it fails to halt.

With these definitions in place, we can prove the Penrose-halting theorem Footnote ³ [Reference Penrose7, pp. 64–66]. Let P be a PSHC, and consider P*, with input p*, its own machine number. If P[p*] is 1 then (i) P*[p*] halts by definition of P, but (ii) by definition of A*, P*[p*] ends in loop and so does not halt; a contradiction. If P[p*] is 0 then (i) P*[p*] does not halt by definition of P, but (ii) by definition of A*, P*[p*] halts on its 0 out arrow; a contradiction. Since P[p*] is neither 1 nor 0, and these are its only possible outputs, for any PSHC P, P[p*] does not halt, hence P*[p*] does not halt either. □

On the basis of (his version of) this theorem Penrose argues (in our terms) against a computationalist theory of cognition: ‘we have a well-defined procedure, whichever [PSHC P] is given to us, for finding a corresponding k for which we know that T $_k(k)$ defeats P, and for which we can therefore do better than the algorithm’ [Reference Penrose7, p. 65]. (By ‘defeats’ he means that P[p*] fails to halt.) That is, we can determine on the basis of the theorem that for all PSHC P, P*[p*] fails to halt. But no PSHC computes for all such P whether P*[p*] halts; in particular, it follows from the theorem that for any PSHC P, P[p*] is undefined—but inputting p* into P is to attempt to determine whether P*[p*] halts. Apparently, then, we can do something that no TM can do!

But this argument is question begging: our ability to see that P*[p*] does not halt is predicated on the information that P is a PSHC, so that the theorem applies. But the PSHC is simply given P*’s machine number p*, and no information that it is the machine number of a PSHC. To compare abilities, both TM and person must be presented with the same task: for instance, given some m—and no other information about T $_m$ —determine whether T $_m$ [m] halts. Of course, that is enough (in principle) to determine concretely which TM in the enumeration T $_m$ is, since it’s computable. Thus to apply the theorem we would have to determine whether T $_m$ is a P* for some PSHC P. Not surprisingly this problem is uncomputable: no general mechanical procedure to determine whether a TM is a PSHC exists, as we shall now show.

We prove that PSHC-hood is not computable by showing that if it were, then there would be a PSHC that can exploit the Penrose-halting theorem to determine that its own P*[p*] halts, contrary to the theorem! That is, we suppose (for reductio) that the following TM exists:

First, recalling that loop is a specific routine, we observe that whether T $_m$ is T $_n^*$ for some T $_n$ is computable: simply determine whether the lone arrow on which T $_m$ halts is of the form

Moreover, since the enumeration of TMs is computable, the value of n is also computable: it is the machine number of the machine obtained from T $_m$ by removing this circuit. Thus the following TM exists:

It thus follows from our supposition that

exists. Call it X.

Suppose that T $_m$ is T $^*_n$ and T $_n$ is a PSHC: by definition the B subroutine outputs $n\neq 0$ , which is input to the S subroutine; and since $T_n$ is a PSHC, S outputs 1 by definition—hence X outputs 0. Otherwise X does not halt: either $T_m$ is not T $^*_n$ for any n, so B outputs 0, and X enters an endless loop; or it is, but T $_n$ is not a PSHC, so S outputs 0 and again X enters an endless loop.

(i) That is, X outputs 1 if T $_m$ is T $^*_n$ and T $_n$ is a PSHC, otherwise it does not halt—X would be a positive solution of the ‘PSHC*-hood problem’.Footnote ⁴

(ii) If X outputs 1 for input m then T $_m[m]$ halts; trivially, since the antecedent is false for any m.

(iii) If X outputs 0 for input m then the S subroutine gave output 1. Thus (a) T $_n$ is a PSHC by definition of S; and (b) the B subroutine output was n, so T $_m$ is T $^*_n$ by definition of B. Hence from the Penrose-halting theorem, T $_m[m]$ does not halt.

(iv) From (ii) and (iii) X is a PSHC.

Let x* be the machine number of X*.

(v) From (i) and (iv), if x* is input into X then the output of X is 0.

(vi) From (iv) and the Penrose-halting theorem, if x* is input into X then X does not halt.

But this is a contradiction, so neither X* nor X exist, so S does not exist (since B does), and contrary to our supposition PSHC-hood is not computable. □

Since according to this corollary to the Penrose-halting theorem no machine has this ability, it is question begging to assume that we possess such an ability, in an argument attempting to show that our cognitive capacities outstrip the computable. But without such an ability we are in no position to use the Penrose-halting theorem to determine that an arbitrary TM A self-halts; to appeal to the theorem one first has to determine that A is P* for some PSHC P, and that in turn first requires that P be identified as a PSHC. So there is no reason to think that our ability to recognize self-halting includes the ability to determine for an arbitrary PSHC P—when not already identified as such—whether P*[p*] halts; no reason to think that we can outperform a TM in this regard.