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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory,sci.logic
Subject: Re: People are still trying to get away with disagreeing with the
 semantics of the x86 language
Date: Mon, 1 Jul 2024 20:22:11 +0200
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Op 01.Jul.2024 OM 17:20 screech Wolcott:
> On 7/1/2024 10:12 AM, Fred. Zwarts wrote:
>> Op 01.jul.2024 om 16:50 schreef olcott:
>>> On 7/1/2024 9:37 AM, Fred. Zwarts wrote:
>>>> Op 01.jul.2024 om 14:46 schreef olcott:
>>>>> On 7/1/2024 3:32 AM, Fred. Zwarts wrote:
>>>>>
>>>>> _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]
>>>>>
>>>>> DDD is correctly emulated by HHH which calls an
>>>>> emulated HHH(DDD) to repeat the process until aborted.
>>>>> Once aborted the DDD emulated by HHH immediately stops.
>>>>>
>>>>> At no point in this emulation does the call from DDD
>>>>> correctly emulated by HHH to HHH(DDD) ever return.
>>>>>
>>>>> You can understand this or fail to understand this
>>>>> disagreement is flat out incorrect.
>>>>
>>>> I understand it, but that does not contradict that the abort is one 
>>>> cycle too soon, which makes it incorrect.
>>>
>>> On 7/1/2024 9:27 AM, Fred. Zwarts wrote:
>>>  > Not aborting will loop infinitely.
>>>
>>> That you disagree with your own self proves that you are wrong.
>>
>> I did not disagree with myself. It is only you inability to understand 
>> simple facts that :
>> It is not: Either aborting or not-aborting is incorrect,
>> but: Both aborting and not-aborting are incorrect.
>> Therefore, proving that not-aborting is incorrect does not prove that 
>> aborting is correct.
>> You never found an error in this reasoning, but only repeat that 
>> not-aborting is incorrect.
>>
>> Somehow you seem to think that from "not-aborting is incorrect" it 
>> follows that "aborting is correct".
>>
>> Try to think a little bit. Both are incorrect.
>>
>>>
>>> If it is ever the case that
>>>  > Not aborting will loop infinitely.
>>> THIS PROVES THAT ABORTING IS NECESSARILY CORRECT
>>>
>>>      If simulating halt decider H correctly simulates its input D
>>>      until H correctly determines that its simulated D would never
>>>      stop running unless aborted then
>>>
>>> IT IS 100% COMPLETELY CORRECT TO ABORT
>>>
>>
>> It must abort to terminate the loop if the simulated HHH would not 
>> terminate, but this simulated HHH does return after N+1 cycles and 
>> therefore DDD will return, so here no abort is needed.
>> Abort is only needed if simulating an infinite recursion, not a 
>> N-cycle recursion.
>>
> 
> You are simply not bright enough to sufficiently understand this
> criteria.
> 
>      If simulating halt decider H correctly simulates its input D
>      until H correctly determines that its simulated D would never
>      stop running unless aborted then

But it does not correctly determine that the simulated D would never 
stop running. Your seem to be incompetent to understand that a C 
program, or x86 language that specifies N cycles of recursive simulation 
does stop after those cycles are all simulated, even when simulated 
correctly.

> 
> void Infinite_Loop()
> {
>    HERE: goto HERE;
> }
> 
> void Infinite_Recursion()
> {
>    Infinite_Recursion();
> }
> 
> void DDD()
> {
>    HHH(DDD);
> }
> 
> int main()
> {
>    HHH(Infinite_Loop);
>    HHH(Infinite_Recursion);
>    HHH(DDD);
> }
> 
> HHH is correct to abort all three.
> 

void Finite_Recursion (nit N) {
   if (N > 0) Finite_Recursion (N - 1);
}

HEH, aborting after N cycles of recursive simulation, when simulating 
itself, is equivalent to Finite_Recursion, not to Infinite_Recursion.