Re: Simulation vs. Execution in the Halting Problem

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:

>
Yes all other people (especially Dennis Bush) are saying
that H(D) is required to report on the behavior of the
direct execution of D() never noticing that this stupidly
requires H(D) to report on the behavior of its caller.

>
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 required by simulation.

>
It turns out that no one ever noticed that simulating halt
deciders nullify the HP counter-example input in that this
input cannot possibly reach its contradictory part.
>

The HP does not care what D does (simply to say).
>

>
Everyone says that H(D) must re[port on the behavior of
the direct execution of D().

>
That is what the HP asks.
>

The HP only requires: H(D)==1 iff D() halts
>
>

>
int main()
{
       D(); // calls H(D)
}
>
Which requires H(D) to report on the behavior of its
caller instead of reporting on the behavior that its
input actually specifies.

>
That is no problem. H does not care what D does inside (simply to say).
The HP simply asks for a H that "H(D)==1 iff D() halts".
>

>
Which requires H to report on something that it cannot possibly see.

>
On the contrary, what the HP proves is very useful.
>

>
I am not talking about the halting problem, I have always
been talking about the conventional halting problem proof.
THIS PROOF IS WRONG

>
When talking about proof, we say it is valid or not. By doing so, we have
to unambiguously pose the problem and the derivation to the conclusion.
The HP proof just did that.
>

>
It may seem that way if you pay less than 100%
complete attention.
>
The HP proof depends on an *INPUT* that does
the opposite of whatever value that H returns
and no such *INPUT* can possibly exist.

 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 undecidable.
 

There is no *input* to any termination analyzer
that can do the opposite of whatever value that
this termination analyzer returns thus the HP
proof itself cannot possibly exist.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer