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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: HHH maps its input to the behavior specified by it --- never
 reaches its halt state
Date: Fri, 9 Aug 2024 08:46:56 +0200
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Op 08.aug.2024 om 21:18 schreef olcott:
> On 8/8/2024 2:04 PM, Fred. Zwarts wrote:
>> Op 08.aug.2024 om 20:48 schreef olcott:
>>> On 8/8/2024 1:44 PM, Fred. Zwarts wrote:
>>>> Op 08.aug.2024 om 15:15 schreef olcott:
>>>>> On 8/8/2024 3:24 AM, Fred. Zwarts wrote:
>>>>>> Op 07.aug.2024 om 15:01 schreef olcott:
>>>>>>> On 8/7/2024 3:16 AM, Fred. Zwarts wrote:
>>>>>>>> Op 04.aug.2024 om 15:11 schreef olcott:
>>>>>>>>> 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?
>>>>>>>>>
>>>>>>>>
>>>>>>>> Is English too difficult for you. I said HHH cannot do it 
>>>>>>>> correctly.
>>>>>>>
>>>>>>> *According to an incorrect criteria of correct*
>>>>>>> You keep trying to get away with disagreeing with
>>>>>>> the semantics of the x86 language. *That is not allowed*
>>>>>>>
>>>>>> Again accusations without evidence.
>>>>>> We proved that HHH deviated from the semantics of the x86 language 
>>>>>> by skipping the last few instructions of a halting program.
>>>>>
>>>>> void DDD()
>>>>> {
>>>>>    HHH(DDD);
>>>>>    return;
>>>>> }
>>>>>
>>>>> Each HHH of every HHH that can possibly exist definitely
>>>>> *emulates zero to infinity instructions correctly* In
>>>>> none of these cases does the emulated DDD ever reach
>>>>> its "return" instruction halt state.
>>>>>
>>>>> *There are no double-talk weasel words around this*
>>>>> *There are no double-talk weasel words around this*
>>>>> *There are no double-talk weasel words around this*
>>>>
>>>> Indeed. And this correctly proves that the simulation failed, not 
>>>> because of an instruction simulated incorrectly, but because 
>>>> instructions are skipped. 
>>>
>>> void Infinite_Recursion()
>>> {
>>>    Infinite_Recursion();
>>>    return;
>>> }
>>
>> Dreaming again of an infinite recursion?
>>
>>>
>>> The return instruction in both cases is unreachable code.
>>> DDD correctly emulated by HHH and Infinite_Recursion
>>> correctly emulated by HHH cannot reach the "return"
>>> instruction.
>>
>> It cannot reach it, because it was programmed to abort one cycle 
>> before the program would end.
>>
> 
> *Maybe you have ADD like Richard has. I already said this above*
> When zero to infinity steps of DDD are correctly emulated by
> HHH no DDD ever reaches its own "return" instruction.
> 
> Maybe the issue is that you don't know programming well enough
> to understand that this is true.

Maybe you should try to learn English. I confirmed hat HHH cannot reach 
the end of the simulation of itself.
Maybe you should learn to program. When the simulation of a halting 
program is unable to reach the end, it proves that the simulation is 
incorrect.
Everybody with sufficient programming knowledge understands that a 
simulator cannot possibly simulate itself correctly up to the end, 
because either it does not halt, or it misses the last cycle, the final 
part of the simulation.