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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 06:08:48 +0800
Organization: A noiseless patient Spider
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In-Reply-To: <105h35a$2uujj$1@dont-email.me>

On Sat, 2025-07-19 at 16:36 -0500, olcott wrote:
> On 7/19/2025 4:26 PM, wij wrote:
> > 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 ca=
n decide
> > > > the halting status of any given D.
> > > >=20
> > >=20
> > > And that is wrong because H(=E2=9F=A8D=E2=9F=A9) is correctly determi=
ned.
> > > It has always been a type mismatch error when H(D) was
> > > assumed.
> >=20
> > Yes, there is type mismatch problems in nearly all discussions.
> > But I don't think you will understand what it is.
> >=20
>=20
> I have proven that I do and you only deny this
> because you are not interested in an honest
> dialogue.

You like to ignore what people say, only insterested in one-sided talk,
showing you are not interested in honest discussion.

> > > > > 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 ha=
lt.
> > > > The conclusion is, no such H exists.
> > > >=20
> > >=20
> > > And that is wrong because H(=E2=9F=A8D=E2=9F=A9) is correctly determi=
ned.
> > > It has always been a type mismatch error when H(D) was
> > > assumed.
> > >=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
> > A type mismatch: HHH(DD) or HHH(<DDD>)?
> >=20
>=20
> DD points to the finite string machine
> description of DD it does not point to
> the executing process of DD.

That is what I predicted: You don't understand what you said.
(because it is a bit technical, I will skip this part)

> > > DD correctly simulated by HHH cannot reach past
> > > the "if" statement thus cannot reach the "return"
> > > statement. T
> >=20
> > That is roughly what HP proof says.
> >=20
>=20
> Not at all. The HP proof claims that DD
> correctly simulated by HHH reaches the
> self-contradictory part of DD and thus
> forms a contradiction.

HP's conclusion is 'undecidable'.

> > > his makes HHH(DD)=3D=3D0 correct.
> >=20
> > How is this statement from?=20
>=20
> You chopped up my statement in the middle of a word.

You jumped to make an arbitrary and contradictory conclusion.
It seems you want to stick your one-sided outcome together with tautology (=
as
usual) hoping it will thus become true. Very ignorant and dishonest.

If I did not chopped up, your whole statement will be a false statement.

> > HHH(DD) above shows it cannot return to report 0.
> > (I guess you might say something and doing another, again)
> >=20
>=20
> Factually incorrect.

No words? If that 'verdict' is all what your 'proof' got? Why bother all th=
e >20
years efforts of 'proof'.

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