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On 7/11/2025 4:07 AM, Fred. Zwarts wrote:No need to repeat it. Everyone sees this failure of HHH to simulate it correctly up to the end.Op 10.jul.2025 om 16:30 schreef olcott:void DDD()On 7/10/2025 6:24 AM, Fred. Zwarts wrote:>Op 09.jul.2025 om 14:54 schreef olcott:>>>One of these "errors" was that HHH cannot simulate itself at all.>
As usual you twist the words of your reviewers.
The claim was that no HHH can simulate itself correctly *up to the end*.
>https://liarparadox.org/HHH(DDD)_Full_Trace.pdf>
Is the full execution trace of
>
executed HHH simulates DDD that calls emulated HHH(DDD)
that simulates DDD that calls emulated emulated HHH(DDD)
And it also proves my claim that HHH did not simulate itself correctly *up to the end*.
>
Exactly what is your professional programming experience?
I have 20 years in C++ and became a professional programmer
in 1986.
>
Irrelevant, even when it is more than your experience.
{
HHH(DDD);
return;
}
Anyone that cannot see that DDD simulated by HHH
cannot possibly reach its own "return" instruction
(a) Has woefully deficient knowledge
(b) Is a liar.
Irrelevant. We know that when it turns out that I have more experience, you will find another irrelevant reason to ignore the facts.Even a proof of a first year student can be a correct proof.I want to see if you have the capacity to understand.
Apparently you ran out of counter arguments and try the authority card.
COBOL experience counts as zero programming experience
relative to anything about recursion.
Yes I do. It means that the halting of DDD depends on the halting of HHH and HHH has conditional branch instructions, which make that it returns to DDD, so that DDD reaches its final halt state, if not disturbed by a premature abort.So, we see that HHH cannot possibly simulate itself up to the end.Because DDD unconditionally calls HHH(DDD).
Do you know what unconditionally means?
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