[FOM] Henk's Characterization of Recursivity
tphhec01 at degas.ceu.hu
Sat Jul 10 13:19:23 EDT 2004
On Thu, Jul 01, 2004 at 04:45:41PM -0400, Ali Enayat wrote:
> Let's begin with the classical characterization of recursivity in terms of
> TM's (Turing Machines):
> 1. A set X is recursive if there is a (deterministic) TM which can "decide"
> all questions of the form "is q in X?". In other words, X is recursive if
> there a TM with the following properties:
> (a) T outputs a single symbol iff T reaches a halting state.
> (b) T either outputs s=0 or s=1 if T reaches the halting state.
> (c) For every input q, there exists a natural number t such that T halts
> after t steps.
> (d) T outputs 1 on input q iff q is in X.
> 2. Some definitions:
> (a) Let us call a TM with properties (a) and (b) a DECISIVE TM. Note that
> (c) and (d) are not included in this definition.
> (b) Suppose T is a decisive TM. Let us say "T accepts q" (T rejects q) if T
> outputs 1 (T outputs 0), when q is the input of T. Also, let us say "T
> decides q" if either T accepts q, or T rejects q.
> 5. Suppose t,e, s, and q are natural numbers. The statement "After t steps,
> the e-th decisive TM outputs s, when provided with q as input" can be
> expressed as a bounded formula. This is definitely classical, and in my
> judgement has been known **at least** since the discovery of Tranhtenbrot's
For making back-reference easy, let's call your bounded formula
psi(t,e,s,q), or if the TM (so its index e) is fixed, psi_e(t,s,q).
> 6. We can now explain the "hard" direction of Henk's characterization:
> Suppose X is a recursive set. Then there is some decisive TM T that
> witnesses the recursivity of X (as in (1) above).
> Consider the formula phi(x) = "the e-th decisive TM accepts x", where T is
> the e-th decisive TM.
What is phi? What I can imagine is that you think of
"there exists t such that psi_e(t,1,x)",
it's a sigma_1 formula.
> Let V be the set of hereditarily finite sets. By (1) through (5), for every
> q, there is some initial segment F(q) of V such that F(q) satisfies "T
> decides q". Therefore the following are equivalent:
This "T decides q" is a bit ambiguous. If we are in V, it's just
equivalent with "True", so then the requirement wrt. F(q) is void.
However, I suppose you'd formalize "T decides q" as follows:
"there exists t such that psi_e(t,0,x) or psi_e(t,1,x)".
> (i) V satisfies phi(q)
> (ii) F(q) satisfies phi(q)
> (iii) Every end extension of F(q) satisfies phi(q).
Now I think you've made mistake here (given that you'd formalize phi in
the way suggested above): (ii) and (iii) are *not*
equivalent. That is, consider the case when q is not in X, ie., T
halts with 0 on q, ie.,
"there exists t such that psi_e(t,0,x)".
In this case, F(q) won't satisfy phi(q), of course.
But it's quite possible that you can end extend F(q) such that phi(q)
will be valid in the extension -- so, in this pathological extension
"there exists t, t' such that psi_e(t,0,x) and psi_e(t',1,x)"
The difficulty lies in re-formulating the above phi in a safe way, which
can cope with those "pathological" extensions. It is possible -- it's
neither trivial nor too hard, doing this doesn't require a genius, one
just have to build the formula carefully, the whole procedure resembles
the way how Rosser extended the scope of Godel's noncompleteness result from
omega-consistent Peano models to arbitrary ones.
Anyway, in my honest opinion the necessity of this re-formulation is
enough for raising my statement above the "folklore" status.
* * *
One more comment: I re-read your proof for the "easy" direction, thanks
for that, it's really an elegant solution, much sorter than my original
"There's more to life, than books, you know but not much more..."
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