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From: wij <wyniijj5@gmail.com>
Newsgroups: comp.software-eng
Subject: Re: Proof that the halting problem itself is a category error
Date: Thu, 11 Dec 2025 07:10:47 +0800
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On Wed, 2025-12-10 at 17:03 -0600, polcott wrote:
> On 12/10/2025 4:58 PM, wij wrote:
> > On Wed, 2025-12-10 at 16:43 -0600, polcott wrote:
> > > When the halting problem requires a halt decider
> > > to report on the behavior of a Turing machine
> > > this is always a category error.
> > >=20
> > > The corrected halting problem requires a Turing
> > > machine decider to report in the behavior that
> > > its finite string input specifies.
> >=20
> > If you honestly admit you are solving POO Problem, everything is fine.
> >=20
>=20
> *It has take me 21 years to boil it down to this*
>=20
> When the halting problem requires a halt decider
> to report on the behavior of a Turing machine this
> is always a category error.
>=20
> The corrected halting problem requires a Turing
> machine decider to report in the behavior that
> its finite string input specifies.

It looks you play blind block my other replies again.
https://sourceforge.net/projects/cscall/files/MisFiles/ghp.txt/download

General Halting Problem, General Undecidable Rule.
-----------
Let MSet be the set of ANY deterministic computing device (including human)=
.
If a device H which computes the decision funcion =E2=88=80x=E2=88=88MSet, =
H(x)=3D1 iff x() halts,
and a device D with contradictory property can be found to exist, then,
contradiction occurres. Both H and D cannot exist in the same set.

    void D() {
      if(H(D)=3D=3D1) for(;;){};  // Both H(D) returning 1 or non-1 contrad=
ict the
    }                         // definition

    int main() {
      H(D);   // H(D) cannot return whith the specified deciding function
    }

    Note: The assertion addresses about any decision funcion, whatever comp=
uting
          the function is irrelevant. The main issue in application of this
          assertion is proving the existence of D in MSet.
    Note: As hinted by Rice's Theorem, if H is about deciding the behaviora=
l
          property of another element in MSet, high possibility is that suc=
h a
          H cannot be an element of MSet. In logic, the answer of propositi=
on
          that both T/F does not fit is referred to as undecidable.

Undecidability may also be so interpreted in the infinite set like the set =
of
natural number: Whenever you have a so-called maximum number n, then n+1 is
automatically defined(generated) to exist in the set to defy the 'maximum'
property... Traditional axiomatized system is limited, no powerful than Tur=
ing
Machine language (or procedural algorithm).
-----------

I have no problem of POO-HP.