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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 07:25:22 +0800
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On Wed, 2025-06-11 at 17:33 -0500, olcott wrote:
> On 6/11/2025 4:57 PM, wij wrote:
> > On Wed, 2025-06-11 at 16:44 -0500, olcott wrote:
> > > On 6/11/2025 4:23 PM, wij wrote:
> > > > 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 saying
> > > > > > > > > > > > > > > that H(D) is required to report on the behavi=
or of the
> > > > > > > > > > > > > > > direct execution of D() never noticing that t=
his stupidly
> > > > > > > > > > > > > > > requires H(D) to report on the behavior of it=
s caller.
> > > > > > > > > > > > > >=20
> > > > > > > > > > > > > > If the H above means the H that the HP refers t=
o. The H is required to
> > > > > > > > > > > > > > report its argument's behavior (ie. by H(D)). B=
ut NOT required by
> > > > > > > > > > > > > > simulation.
> > > > > > > > > > > > >=20
> > > > > > > > > > > > > It turns out that no one ever noticed that simula=
ting halt
> > > > > > > > > > > > > deciders nullify the HP counter-example input in =
that this
> > > > > > > > > > > > > input cannot possibly reach its contradictory par=
t.
> > > > > > > > > > > > >=20
> > > > > > > > > > > > > > The HP does not care what D does (simply to say=
).
> > > > > > > > > > > > > >=20
> > > > > > > > > > > > >=20
> > > > > > > > > > > > > Everyone says that H(D) must re[port on the behav=
ior 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=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 ins=
ide (simply 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 canno=
t possibly 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 doi=
ng so, we have
> > > > > > > > to unambiguously pose the problem and the derivation to the=
 conclusion.
> > > > > > > > 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 possib=
le 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 u=
ndecidable.
> > > > > >=20
> > > > >=20
> > > > > There is no *input* to any termination analyzer
> > > > > that can do the opposite of whatever value that
> > > > > this termination analyzer returns
> > > >=20
> > > > Your reinterpretation of of HP case is wrong.
> > > > Your D or H is not the case mention in the HP proof.
> > > >=20
> > >=20
> > > There cannot possibly exist any D mine or
> > > anyone else's that is encoded to do the opposite
> > > of whatever value that H returns.
> >=20
> > Why not? D and H are supposed to be TM (or C function).
> > If the D cannot do the opposite of whatever value that H returns, then
> > that D is not powerful enough to be a TM, not an interesting case.
> >=20
>=20
> Can you be your biological mother's biological father?

What is the same reason? What's the relationship of 1+1=3D2 relates to HP?

> It is for this same reason that the function's caller
> cannot simultaneously be its input.

D and H belong to the same set of TM equivalent stuff.
D has to be able to perform exactly H's function (if D is a TM and if H exi=
sts).
Otherwise, that D is not the counter-example mentioned in the HP proof.