Re: Analysis of Flibble’s Latest: Detecting vs. Simulating Infinite Recursion ZFC

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Sujet : Re: Analysis of Flibble’s Latest: Detecting vs. Simulating Infinite Recursion ZFC
De : F.Zwarts (at) *nospam* HetNet.nl (Fred. Zwarts)
Groupes : comp.theory
Date : 27. May 2025, 09:56:54
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <1013uon$2h8vk$2@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
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Op 27.mei.2025 om 10:48 schreef Fred. Zwarts:
Op 26.mei.2025 om 22:28 schreef Richard Heathfield:
On 26/05/2025 20:35, Fred. Zwarts wrote:
Now DDD is not the caller of HHH
>
Yes, it is.
>
void DDD()
{
   HHH(DDD);
   return;
}
>
 Only if we confuse levels of simulation.
    int main () {
    HHH(DDD);
  }
 Here main is the caller of HHH. In the first level of simulation DDD is simulated and calls HHH, which could cause a second level of simulation, where DDD is the caller, but that does not happen, because the first HHH aborts the simulation at that point. At the first (and only level), DDD is not the caller of HHH. DDD is only the input to HHH, not the caller. HHH should decide about its input, not about its caller.
DDD is, in fact, a pointer to memory. This memory includes the code of DDD and all other code used by DDD, including the HHH that aborts. So, the input specifies a halting program. But HHH does not see that part of the specification, but aborts and makes the false assumption about itself that it does not halt. This bug in HHH does not change the verifiable fact that the input (not the caller) specifies a halting program.
That Olcott confuses himself by presenting only a DDD that calls HHH, does not change the fact that HHH should decide about its input, even if the code of the caller is included in the input. HHH should decide about its input, not about its caller. That the code of the caller happens to be the same as the code that is included in the input is a completely irrelevant detail.

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