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From: wij <wyniijj5@gmail.com>
Newsgroups: comp.ai.philosophy
Subject: Re: I have just proven the error of all of the halting problem
 proofs --- Mackenzie
Date: Sun, 27 Jul 2025 11:08:20 +0800
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On Sat, 2025-07-26 at 21:43 -0500, olcott wrote:
> On 7/26/2025 8:30 PM, Richard Damon wrote:
> > On 7/26/25 7:43 PM, olcott wrote:
> > > On 7/26/2025 6:35 PM, Richard Damon wrote:
> > > > On 7/26/25 7:08 PM, olcott wrote:
> > > > > On 7/26/2025 5:49 PM, olcott wrote:
> > > > > > On 7/26/2025 2:58 PM, olcott wrote:
> > > > > > > On 7/26/2025 2:52 PM, Mr Flibble wrote:
> > > > > > > > On Sat, 26 Jul 2025 14:26:27 -0500, olcott wrote:
> > > > > > > >=20
> > > > > > > > > On 7/26/2025 1:30 PM, Alan Mackenzie wrote:
> > > > > > > > > > In comp.theory olcott <polcott333@gmail.com> wrote:
> > > > > > > > > >=20
> > > > > > > > > > > The error of all of the halting problem proofs is tha=
t they=20
> > > > > > > > > > > require a
> > > > > > > > > > > Turing machine halt decider to report on the behavior=
 of a=20
> > > > > > > > > > > directly
> > > > > > > > > > > executed Turing machine.
> > > > > > > > > >=20
> > > > > > > > > > > It is common knowledge that no Turing machine decider=
 can take=20
> > > > > > > > > > > another
> > > > > > > > > > > directly executing Turing machine as an input, thus t=
he above
> > > > > > > > > > > requirement is not precisely correct.
> > > > > > > > > >=20
> > > > > > > > > > > When we correct the error of this incorrect requireme=
nt it=20
> > > > > > > > > > > becomes a
> > > > > > > > > > > Turing machine decider indirectly reports on the beha=
vior of a
> > > > > > > > > > > directly executing Turing machine through the proxy o=
f a=20
> > > > > > > > > > > finite string
> > > > > > > > > > > description of this machine.
> > > > > > > > > >=20
> > > > > > > > > > > Now I have proven and corrected the error of all of t=
he halting
> > > > > > > > > > > problem proofs.
> > > > > > > > > >=20
> > > > > > > > > > No you haven't, the subject matter is too far beyond yo=
ur=20
> > > > > > > > > > intellectual
> > > > > > > > > > capacity.
> > > > > > > > > >=20
> > > > > > > > > >=20
> > > > > > > > > It only seems to you that I lack understanding because yo=
u are=20
> > > > > > > > > so sure
> > > > > > > > > that I must be wrong that you make sure to totally ignore=
 the=20
> > > > > > > > > subtle
> > > > > > > > > nuances of meaning that proves I am correct.
> > > > > > > > >=20
> > > > > > > > > No Turing machine based (at least partial) halt decider c=
an=20
> > > > > > > > > possibly
> > > > > > > > > *directly* report on the behavior of any directly executi=
ng Turing
> > > > > > > > > machine.=C2=A0 The best that any of them can possibly do =
is=20
> > > > > > > > > indirectly report
> > > > > > > > > on this behavior through the proxy of a finite string mac=
hine
> > > > > > > > > description.
> > > > > > > >=20
> > > > > > > > Partial decidability is not a hard problem.
> > > > > > > >=20
> > > > > > > > /Flibble
> > > > > > >=20
> > > > > > > My point is that all of the halting problem proofs
> > > > > > > are wrong when they require a Turing machine decider
> > > > > > > H to report on the behavior of machine M on input i
> > > > > > > because machine M is not in the domain of any Turing
> > > > > > > machine decider. Only finite strings such as =E2=9F=A8M=E2=9F=
=A9 the
> > > > > > > Turing machine description of machine M are its
> > > > > > > domain.
> > > > > > >=20
> > > > > >=20
> > > > > > Definition of Turing Machine =C4=A4
> > > > > > =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 halt=
s, and=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 // incorrect requirement
> > > > > > =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.=C2=A0=C2=A0=C2=A0 // incorrect requirement
> > > > > >=20
> > > > > > (a) =C4=A4 copies its input =E2=9F=A8=C4=A4=E2=9F=A9
> > > > > > (b) =C4=A4 invokes embedded_H =E2=9F=A8=C4=A4=E2=9F=A9 =E2=9F=
=A8=C4=A4=E2=9F=A9
> > > > > > (c) embedded_H simulates =E2=9F=A8=C4=A4=E2=9F=A9 =E2=9F=A8=C4=
=A4=E2=9F=A9
> > > > > > (d) simulated =E2=9F=A8=C4=A4=E2=9F=A9 copies its input =E2=9F=
=A8=C4=A4=E2=9F=A9
> > > > > > (e) simulated =E2=9F=A8=C4=A4=E2=9F=A9 invokes simulated embedd=
ed_H =E2=9F=A8=C4=A4=E2=9F=A9 =E2=9F=A8=C4=A4=E2=9F=A9
> > > > > > (f) simulated embedded_H simulates =E2=9F=A8=C4=A4=E2=9F=A9 =E2=
=9F=A8=C4=A4=E2=9F=A9 ...
> > > > > >=20
> > > > > > The fact that the correctly simulated input
> > > > > > specifies recursive simulation prevents the
> > > > > > simulated =E2=9F=A8=C4=A4=E2=9F=A9 from ever reaching its simul=
ated
> > > > > > final halt state of =E2=9F=A8=C4=A4.qn=E2=9F=A9, thus specifies=
 non-termination.
> > > > > >=20
> > > > > > This is not contradicted by the fact that
> > > > > > =C4=A4 applied to =E2=9F=A8=C4=A4=E2=9F=A9 halts because =C4=A4=
 is outside of
> > > > > > the domain of every Turing machine computed function.
> > > > > >=20
> > > > >=20
> > > > > In the atypical case where the behavior of the simulation
> > > > > of an input to a potential halt decider disagrees with the
> > > > > behavior of the direct execution of the underlying machine
> > > > > (because this input calls this same simulating decider) it
> > > > > is the behavior of the input that rules because deciders
> > > > > compute the mapping for their inputs.
> > > > >=20
> > > >=20
> > > > Nope, just more of your lies.
> > > >=20
> > > > The behavior of an input to a halt decider is DEFINED in all cases =
to=20
> > > > be the behavior of the machine the input represents,
> > >=20
> > > Yet I have conclusively proven otherwise and
> > > you are too stupid to understand the proof.
> >=20
> > No, because you proof needs to call different inputs the same or partia=
l=20
> > simulaiton to be correct.
> >=20
>=20
> When HHH(DDD) simulates DDD it also simulates itself
> simulating DDD because DDD calls HHH(DDD).
>=20
> When HHH1(DDD) simulates DDD DOES NOT simulate itself
> simulating DDD because DDD DOES NOT CALL HHH1(DDD).
>=20
> For three fucking years everyone here pretended that
> they could NOT fucking see that.

It is you who proved yourself an idiot, worse, a liar, EVERYDAY.

olcott's claim form H(D)=3D0 is correct, H(D)=3D1 is correct, both are corr=
ect...
'I' was talking about HH,HH2,HHH, DD,DDD,...not H(D)!!  ... numerous.
And recently, 'I' was not refuting HP. HP is correct. 'I' was refuting Linz=
's
proof, and HHH(DD)=3D1 is correct!! (Undecidable and HHH(DD)=3D1 are both c=
orrect!)

A couple days before, you have shown again you don't understand basic logic=
 (AND,IF,...)
You cannot construct a TM that compute the length of its input.
Your understanding of C/Assembly is shown very low, I never saw anyone is l=
ower.
No one in internet I ever saw is lower than yours. Keep blind yourself, 'ge=
nius'.