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Op 09.jun.2025 om 16:39 schreef olcott:It waiting forever is not long enough (and it is)On 6/9/2025 5:26 AM, Fred. Zwarts wrote:It is even more stupid to ignore the halting part of the input (with a premature abort) and claim it is not halting.Op 09.jun.2025 om 06:04 schreef olcott:>On 6/8/2025 10:54 PM, Keith Thompson wrote:>olcott <polcott333@gmail.com> writes:>On 6/8/2025 10:31 PM, Keith Thompson wrote:>olcott <polcott333@gmail.com> writes:>void DDD()Do not imply that I support your claims.
{
HHH(DDD);
return;
}
>
The *input* to simulating termination analyzer HHH(DDD)
specifies recursive simulation that can never reach its
*simulated "return" instruction final halt state*
>
*Every rebuttal to this changes the words*
I am not implying anything. I am directly stating
that you have agreed that when DDD is correctly simulated
by HHH that it cannot possibly reach its own simulated
"return" instruction and terminate normally.
Endless recursion is endless recursion. Correctly simulated endless
recursion is endless recursion.
Great. No one else besides you and I agree that DDD
correctly simulated by HHH cannot possibly reach its
*simulated "return" instruction final halt state*
Nobody denied it. You are fighting windmills.
We all agree that your HHH fails to reach the end of the simulation of the input. An input that specifies a halting program, but HHH cannot simulate it.
>>>This has no useful or interesting>
consequences. Do you agree?
>
It is very useful because it is isomorphic to this:
(The standard Halting Problem counter-example input)
>
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
>
Indeed, it shows that simulation is not the right way to try to refute the proof of the halting theorem, because a simulator will never be able to simulate itself correctly up to the end.
>
It is ridiculously stupid to require a non-terminating
input to be simulated up to its non-existent end.
>
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