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From: wij <wyniijj5@gmail.com>
Newsgroups: comp.theory
Subject: Re: Four Chatbots figure out on their own without prompting that
 HHH(DDD)==0
Date: Sun, 20 Jul 2025 04:14:05 +0800
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On Sat, 2025-07-19 at 15:01 -0500, olcott wrote:
> On 7/19/2025 2:59 PM, wij wrote:
> > On Sat, 2025-07-19 at 14:47 -0500, olcott wrote:
> > > On 7/19/2025 2:29 PM, wij wrote:
> > > > On Sat, 2025-07-19 at 14:19 -0500, olcott wrote:
> > > > > On 7/19/2025 12:02 PM, Richard Damon wrote:
> > > > > > On 7/19/25 10:42 AM, olcott wrote:
> > > > > > > On 7/18/2025 3:49 AM, joes wrote:
> > > > > > >=20
> > > > > > > > That is wrong. It is, as you say, very obvious that HHH can=
not simulate
> > > > > > > > DDD past the call to HHH. You just draw the wrong conclusio=
n from it.
> > > > > > > > (Aside: what "seems" to you will convince no one. You can j=
ust call
> > > > > > > > everybody dishonest. Also, they are not "your reviewers".)
> > > > > > > >=20
> > > > > > >=20
> > > > > > > For the purposes of this discussion this is the
> > > > > > > 100% complete definition of HHH. It is the exact
> > > > > > > same one that I give to all the chat bots.
> > > > > > >=20
> > > > > > > Termination Analyzer HHH simulates its input until
> > > > > > > it detects a non-terminating behavior pattern. When
> > > > > > > HHH detects such a pattern it aborts its simulation
> > > > > > > and returns 0.
> > > > > >=20
> > > > > > So, the only HHH that meets your definition is the HHH that nev=
er
> > > > > > detects the pattern and aborts, and thus never returns.
> > > > > >=20
> > > > >=20
> > > > > All of the Chat bots conclude that HHH(DDD) is correct
> > > > > to reject its input as non-halting because this input
> > > > > specified recursive simulation. They figure this out
> > > > > on their own without any prompting.
> > > > >=20
> > > > > https://chatgpt.com/share/687aa4c2-b814-8011-9e7d-b85c03b291eb
> > > >=20
> > > > It is still nothing to do with the Halting Problem proof (Because i=
t is POOH)
> > > >=20
> > >=20
> > > It is a key element of my refutation of this proof
> > > because HHH also correctly determines that HHH(DD)=3D=3D0.
> > >=20
> > > DD correctly simulated by HHH cannot possibly ever
> > > reach past its first statement because it specifies
> > > recursive simulation.
> > >=20
> > > int DD()
> > > {
> > > =C2=A0=C2=A0=C2=A0 int Halt_Status =3D HHH(DD);
> > > =C2=A0=C2=A0=C2=A0 if (Halt_Status)
> > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 HERE: goto HERE;
> > > =C2=A0=C2=A0=C2=A0 return Halt_Status;
> > > }
> > >=20
> >=20
> > Boring. HHH cannot do what the HP says.
> >=20
>=20
> Turing machine (at least partial) halt deciders only compute
> the mapping from their finite string inputs to the actual
> behavior that this input finite string actually specifies.
>=20
> Conventional notation of a Turing Machine: =C4=A4
> Conventional notation of a TM description: =E2=9F=A8=C4=A4=E2=9F=A9
>=20
> =C4=A4.q0 =E2=9F=A8=C4=A4=E2=9F=A9 =E2=8A=A2* =C4=A4.embedded_H =E2=9F=A8=
=C4=A4=E2=9F=A9 =E2=9F=A8=C4=A4=E2=9F=A9 =E2=8A=A2* =C4=A4.=E2=88=9E,
> =C2=A0=C2=A0 if =C4=A4 applied to =E2=9F=A8=C4=A4=E2=9F=A9 halts, and
> =C4=A4.q0 =E2=9F=A8=C4=A4=E2=9F=A9 =E2=8A=A2* =C4=A4.embedded_H =E2=9F=A8=
=C4=A4=E2=9F=A9 =E2=9F=A8=C4=A4=E2=9F=A9 =E2=8A=A2* =C4=A4.qn
> =C2=A0=C2=A0 if =C4=A4 applied to =E2=9F=A8=C4=A4=E2=9F=A9 does not halt.
>=20
> *Is corrected to this*
>=20
> =C4=A4.q0 =E2=9F=A8=C4=A4=E2=9F=A9 =E2=8A=A2* =C4=A4.embedded_H =E2=9F=A8=
=C4=A4=E2=9F=A9 =E2=9F=A8=C4=A4=E2=9F=A9 =E2=8A=A2* =C4=A4.=E2=88=9E
> =C2=A0=C2=A0=C2=A0=C2=A0 =E2=9F=A8=C4=A4=E2=9F=A9 =E2=9F=A8=C4=A4=E2=9F=
=A9 simulated by =C4=A4.embedded_H reaches
> =C2=A0=C2=A0=C2=A0=C2=A0 its simulated final halt state of =E2=9F=A8=C4=
=A4.qn=E2=9F=A9, and
> =C4=A4.q0 =E2=9F=A8=C4=A4=E2=9F=A9 =E2=8A=A2* =C4=A4.embedded_H =E2=9F=A8=
=C4=A4=E2=9F=A9 =E2=9F=A8=C4=A4=E2=9F=A9 =E2=8A=A2* =C4=A4.qn
> =C2=A0=C2=A0=C2=A0=C2=A0 =E2=9F=A8=C4=A4=E2=9F=A9 =E2=9F=A8=C4=A4=E2=9F=
=A9 simulated by =C4=A4.embedded_H cannot possibly
> =C2=A0=C2=A0=C2=A0=C2=A0 reach its simulated final halt state of =E2=9F=
=A8=C4=A4.qn=E2=9F=A9.
>=20
> *Original proof*
> https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

https://en.wikipedia.org/wiki/Halting_problem
https://brilliant.org/wiki/halting-problem/
https://www.geeksforgeeks.org/theory-of-computation/halting-problem-in-theo=
ry-of-computation/
https://www.sciencedirect.com/science/article/pii/S235222082100050X

Modifying historical fact is nut.

HP is very simple: H(D)=3D1 if D halts, H(D)=3D0 if D does not halt.