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From: olcott <abc@def.com>
Newsgroups: comp.theory
Subject: Re: Who here is too stupid to know that DDD correctly simulated by
 HHH cannot possibly reach its own return instruction?
Date: Wed, 7 Aug 2024 08:40:31 -0500
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On 8/7/2024 2:22 AM, Mikko wrote:
> On 2024-08-05 15:00:12 +0000, olcott said:
> 
>> On 8/5/2024 2:44 AM, Mikko wrote:
>>> On 2024-08-04 13:11:56 +0000, olcott said:
>>>
>>>> On 8/4/2024 1:26 AM, Fred. Zwarts wrote:
>>>>> Op 03.aug.2024 om 17:20 schreef olcott:>>
>>>>>> When you try to show how DDD simulated by HHH does
>>>>>> reach its "return" instruction you must necessarily
>>>>>> must fail unless you cheat by disagreeing with the
>>>>>> semantics of C. That you fail to have a sufficient
>>>>>> understanding of the semantics of C is less than no
>>>>>> rebuttal what-so-ever.
>>>>>
>>>>> Fortunately that is not what I try, because I understand that HHH 
>>>>> cannot possibly simulate itself correctly.
>>>>>
>>>>
>>>> void DDD()
>>>> {
>>>>    HHH(DDD);
>>>>    return;
>>>> }
>>>>
>>>> In other words when HHH simulates itself simulating DDD it
>>>> is supposed to do something other than simulating itself
>>>> simulating DDD ???  Do you expect it to make a cup of coffee?
>>>
>>> In another message you have said that when HHH simulates itself
>>> simulating DDD is does not simulate itself simulating itself
>>> simulating DDD. You have not told whether it makes a cup of coffee.
>>> Neither action can be seen in the traces you have shown.
>>>
>>
>> HHH and HH and the original H have proved that they simulate
>> themselves simulating DDD, DD and P for three years now.
> 
> Your trace don't show siulation of exectuion differently from
> simulation of simulation of execution.
> 

It does but it is too difficult to dig it out of emulations
of emulators emulating inputs.

You can go in there and highlight all of the instructions
of the first emulated DDD in yellow, the second DDD in cyan,
the first HHH in red and the second HHH in green.

As soon as the first HHH sees the second DDD about to invoke
a third HHH it aborts the emulation. At this point DDD, the
second HHH and the second DDD all immediately stop running
and HHH returns 0 to main.

https://liarparadox.org/HHH(DDD)_Full_Trace.pdf

void DDD()
{
   HHH(DDD);
   return;
}

No actual need to go through all that tediousness. Any expert
in the C language that knows what x86 emulators are knows
the DDD correctly emulated by HHH specifies what is essentially
equivalent to infinite recursion. It seems that no one here
has that degree of expertise.

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer