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From: wij <wyniijj5@gmail.com>
Newsgroups: comp.ai.philosophy
Subject: Re: Four Chatbots figure out on their own without prompting that
 HHH(DDD)==0
Date: Sun, 20 Jul 2025 05:26:34 +0800
Organization: A noiseless patient Spider
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In-Reply-To: <105h1ar$2uj5e$1@dont-email.me>

On Sat, 2025-07-19 at 16:05 -0500, olcott wrote:
> On 7/19/2025 3:57 PM, wij wrote:
> > On Sat, 2025-07-19 at 15:41 -0500, olcott wrote:
> > > On 7/19/2025 3:14 PM, wij wrote:
> > > >=20
> > > > HP is very simple: H(D)=3D1 if D halts, H(D)=3D0 if D does not halt=
..
> > > >=20
> > >=20
> > > The standard proof assumes a decider
> > > H(M,x) that determines whether machine
> > > M halts on input x.
> > >=20
> > > But this formulation is flawed, because:
> >=20
> > Whatever the 'formulation' is, the HP result is a fact that no H can de=
cide
> > the halting status of any given D.
> >=20
>=20
> And that is wrong because H(=E2=9F=A8D=E2=9F=A9) is correctly determined.
> It has always been a type mismatch error when H(D) was
> assumed.

Yes, there is type mismatch problems in nearly all discussions.
But I don't think you will understand what it is.

> > > Turing machines can only process finite encodings
> > > (e.g. =E2=9F=A8M=E2=9F=A9), not executable entities like M.
> > >=20
> > > So the valid formulation must be
> > > H(=E2=9F=A8M=E2=9F=A9,x), where =E2=9F=A8M=E2=9F=A9 is a string.
> >=20
> > Halting Problem::=3D H(D)=3D1 if D halts, H(D)=3D0 if D does not halt.
> > The conclusion is, no such H exists.
> >=20
>=20
> And that is wrong because H(=E2=9F=A8D=E2=9F=A9) is correctly determined.
> It has always been a type mismatch error when H(D) was
> assumed.
>=20
> int DD()
> {
> =C2=A0=C2=A0 int Halt_Status =3D HHH(DD);
> =C2=A0=C2=A0 if (Halt_Status)
> =C2=A0=C2=A0=C2=A0=C2=A0 HERE: goto HERE;
> =C2=A0=C2=A0 return Halt_Status;
> }
>=20

A type mismatch: HHH(DD) or HHH(<DDD>)?

> DD correctly simulated by HHH cannot reach past
> the "if" statement thus cannot reach the "return"
> statement. T

That is roughly what HP proof says.

> his makes HHH(DD)=3D=3D0 correct.

How is this statement from? HHH(DD) above shows it cannot return to report =
0.
(I guess you might say something and doing another, again)

> > 'formulation' does not really matter.
> > If 'formulation' matters, it is another problem.
> >=20
>=20
>=20