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From: wij <wyniijj5@gmail.com>
Newsgroups: comp.theory
Subject: Re: Simulation vs. Execution in the Halting Problem
Date: Thu, 12 Jun 2025 05:23:07 +0800
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In-Reply-To: <102crbv$26rt0$1@dont-email.me>

On Wed, 2025-06-11 at 16:10 -0500, olcott wrote:
> On 6/11/2025 3:59 PM, wij wrote:
> > On Wed, 2025-06-11 at 15:30 -0500, olcott wrote:
> > > On 6/11/2025 2:45 PM, wij wrote:
> > > > On Wed, 2025-06-11 at 14:39 -0500, olcott wrote:
> > > > > On 6/11/2025 2:31 PM, wij wrote:
> > > > > > On Wed, 2025-06-11 at 14:14 -0500, olcott wrote:
> > > > > > > On 6/11/2025 1:25 PM, wij wrote:
> > > > > > > > On Wed, 2025-06-11 at 12:59 -0500, olcott wrote:
> > > > > > > > > > >=20
> > > > > > > > > > > Yes all other people (especially Dennis Bush) are say=
ing
> > > > > > > > > > > that H(D) is required to report on the behavior of th=
e
> > > > > > > > > > > direct execution of D() never noticing that this stup=
idly
> > > > > > > > > > > requires H(D) to report on the behavior of its caller=
..
> > > > > > > > > >=20
> > > > > > > > > > If the H above means the H that the HP refers to. The H=
 is required to
> > > > > > > > > > report its argument's behavior (ie. by H(D)). But NOT r=
equired by simulation.
> > > > > > > > >=20
> > > > > > > > > It turns out that no one ever noticed that simulating hal=
t
> > > > > > > > > deciders nullify the HP counter-example input in that thi=
s
> > > > > > > > > input cannot possibly reach its contradictory part.
> > > > > > > > >=20
> > > > > > > > > > The HP does not care what D does (simply to say).
> > > > > > > > > >=20
> > > > > > > > >=20
> > > > > > > > > Everyone says that H(D) must re[port on the behavior of
> > > > > > > > > the direct execution of D().
> > > > > > > >=20
> > > > > > > > That is what the HP asks.
> > > > > > > >=20
> > > > > > > > > > The HP only requires: H(D)=3D=3D1 iff D() halts
> > > > > > > > > >=20
> > > > > > > > > >=20
> > > > > > > > >=20
> > > > > > > > > int main()
> > > > > > > > > {
> > > > > > > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 D(); // calls H(D)
> > > > > > > > > }
> > > > > > > > >=20
> > > > > > > > > Which requires H(D) to report on the behavior of its
> > > > > > > > > caller instead of reporting on the behavior that its
> > > > > > > > > input actually specifies.
> > > > > > > >=20
> > > > > > > > That is no problem. H does not care what D does inside (sim=
ply to say).
> > > > > > > > The HP simply asks for a H that "H(D)=3D=3D1 iff D() halts"=
..
> > > > > > > >=20
> > > > > > >=20
> > > > > > > Which requires H to report on something that it cannot possib=
ly see.
> > > > > >=20
> > > > > > On the contrary, what the HP proves is very useful.
> > > > > >=20
> > > > >=20
> > > > > I am not talking about the halting problem, I have always
> > > > > been talking about the conventional halting problem proof.
> > > > > THIS PROOF IS WRONG
> > > >=20
> > > > When talking about proof, we say it is valid or not. By doing so, w=
e have
> > > > to unambiguously pose the problem and the derivation to the conclus=
ion.
> > > > The HP proof just did that.
> > > >=20
> > >=20
> > > It may seem that way if you pay less than 100%
> > > complete attention.
> > >=20
> > > The HP proof depends on an *INPUT* that does
> > > the opposite of whatever value that H returns
> > > and no such *INPUT* can possibly exist.
> >=20
> > That is absolutely correct. No such *INPUT* (i.e. D) can possible exit =
is because
> > the H inside D does not exist at all.
> > So, if the H is assumed to exist, then D will exist to make H undecidab=
le.
> >=20
>=20
> There is no *input* to any termination analyzer
> that can do the opposite of whatever value that
> this termination analyzer returns=C2=A0

Your reinterpretation of of HP case is wrong. Your D or H is not the case m=
ention in the HP proof.

H and D must be in the same set to talk about HP problem.
H and D must be both "C/Assembly" functions, or executables, or OBJ files (=
if you like)...etc.

> thus the HP
> proof itself cannot possibly exist.