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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: =?UTF-8?Q?Re=3A_Analysis_of_Flibble=E2=80=99s_Latest=3A_Detecting_v?=
 =?UTF-8?Q?s=2E_Simulating_Infinite_Recursion_ZFC?=
Date: Tue, 27 May 2025 10:48:57 +0200
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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.