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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: DDD emulated by HHH --- (does not refer to prior posts)
Date: Thu, 29 Aug 2024 19:50:25 +0200
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Op 29.aug.2024 om 19:32 schreef olcott:
> On 8/29/2024 12:22 PM, Fred. Zwarts wrote:
>> Op 29.aug.2024 om 16:07 schreef olcott:
>>> On 8/29/2024 2:17 AM, Mikko wrote:
>>>> On 2024-08-28 12:08:06 +0000, olcott said:
>>>>
>>>>> On 8/28/2024 2:39 AM, Mikko wrote:
>>>>>> On 2024-08-27 12:44:31 +0000, olcott said:
>>>>>>
>>>>>>> On 8/27/2024 3:38 AM, Fred. Zwarts wrote:
>>>>>>>> Op 27.aug.2024 om 04:33 schreef olcott:
>>>>>>>>> This is intended to be a stand-alone post that does not
>>>>>>>>> reference anything else mentioned in any other posts.
>>>>>>>>>
>>>>>>>>> void DDD()
>>>>>>>>> {
>>>>>>>>>    HHH(DDD);
>>>>>>>>>    return;
>>>>>>>>> }
>>>>>>>>>
>>>>>>>>> _DDD()
>>>>>>>>> [00002172] 55         push ebp      ; housekeeping
>>>>>>>>> [00002173] 8bec       mov ebp,esp   ; housekeeping
>>>>>>>>> [00002175] 6872210000 push 00002172 ; push DDD
>>>>>>>>> [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
>>>>>>>>> [0000217f] 83c404     add esp,+04
>>>>>>>>> [00002182] 5d         pop ebp
>>>>>>>>> [00002183] c3         ret
>>>>>>>>> Size in bytes:(0018) [00002183]
>>>>>>>>>
>>>>>>>>> When we assume that:
>>>>>>>>> (a) HHH is an x86 emulator that is in the same memory space as 
>>>>>>>>> DDD.
>>>>>>>>> (b) HHH emulates DDD according to the semantics of the x86 
>>>>>>>>> language.
>>>>>>>>>
>>>>>>>>> then we can see that DDD emulated by HHH cannot possibly get past
>>>>>>>>> its own machine address 0000217a.
>>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>> Yes, we see. In fact DDD is not needed at all.
>>>>>>>
>>>>>>> A straw man fallacy (sometimes written as strawman) is the 
>>>>>>> informal fallacy of refuting an argument different from the one 
>>>>>>> actually under discussion...
>>>>>>> https://en.wikipedia.org/wiki/Straw_man
>>>>>>
>>>>>> You should also point a link to the equivocation fallacy. You use it
>>>>>> more often than straw man.
>>>>>
>>>>> Isomorphism is not equivocation
>>>>
>>>> The use of HHH for many purposes (a specific program, an unpsecified
>>>> memeber of a set of programs, a hypothetical program) is.
>>>>
>>>> Your first posting looked like you were going to apply equivocation
>>>> later in the discussion. Now, after several later messages, it seems
>>>> that you want to apply the fallacy of "moving the goal posts" instead.
>>>>
>>>
>>> void EEE()
>>> {
>>>    HERE: goto HERE;
>>>    return;
>>> }
>>>
>>> HHH correctly predicts what the behavior of EEE would
>>> be if this HHH never aborted its emulation of EEE.
>>>
>>> void DDD()
>>> {
>>>    HHH(DDD);
>>>    return;
>>> }
>>>
>>> HHH correctly predicts what the behavior of DDD would
>>> be if this HHH never aborted its emulation of DDD.
>> Which is incorrect, because HHH is not allowed to change the input. 
>> The simulating HHH may abort, but it may not ignore the fact that the 
>> input (the simulated HHH) is coded to abort when it sees the 'special 
>> condition'. Otherwise it would decide about a non-input, which is not 
>> allowed.
>>
> 
> *I told you this too many times so you must be a liar*
> No DDD ever reaches its "return" instruction no matter
> what-the-Hell that HHH does, 

Exactly. Do you finally understand that HHH cannot possibly simulate 
itself up to the end?
No matter what you do or say, HHH cannot possibly simulate itself 
correctly up to the end. That is the reason, why HHH does not return to 
DDD.
You are such a slow learner. In how many different words do I have to 
repeat this before you finally understand what I say?

thus DDD CANNOT POSSIBLY HALT.
> 

DDD does not halt because HHH does not allow it to halt, because the 
simulating HHH aborted the simulated HHH one cycle before it would halt. 
The simulation stopped prematurely.
A correct simulation (like the one by HHH1) shows that DDD *does* halt.
This proves that "DDD CANNOT POSSIBLY HALT" is simply not true. It can 
be shown in many ways the DDD *does* halt.
The semantics of the X86 language do not change when the simulator is 
called HHH. A halting program will always be a halting program according 
to the semantics of the x86 language.
I told you so many times that HHH cannot possibly simulate itself 
correctly up to the end. That is the problem, not DDD.

        int main() {
          return HHH(main);
        }

HHH halts, but decides that it does not halt. It is HHH that is 
self-contradictory.