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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory,sci.logic
Subject: Re: Flat out dishonest or totally ignorant? --- Richard seems to be
 willfully ignorant
Date: Wed, 3 Jul 2024 20:04:29 +0200
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Op 03.jul.2024 om 18:11 schreef olcott:
> On 7/3/2024 11:05 AM, Fred. Zwarts wrote:
>> Op 03.jul.2024 om 17:53 schreef olcott:
>>> On 7/3/2024 10:44 AM, Fred. Zwarts wrote:
>>>> Op 03.jul.2024 om 15:22 schreef olcott:
>>>>> On 7/3/2024 3:33 AM, Fred. Zwarts wrote:
>>>>>> Op 03.jul.2024 om 05:07 schreef olcott:
>>>>>>> On 7/2/2024 9:35 PM, Richard Damon wrote:
>>>>>>>> On 7/2/24 10:03 PM, olcott wrote:
>>>>>>>>> On 7/2/2024 8:51 PM, Richard Damon wrote:
>>>>>>>>>> On 7/2/24 9:32 PM, olcott wrote:
>>>>>>>>>>> On 7/2/2024 8:25 PM, Richard Damon wrote:
>>>>>>>>>>>> On 7/2/24 9:18 PM, olcott wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> Professor Sipser probably does understand the x86 language.
>>>>>>>>>>>>> Shared-memory implementation of the Karp-Sipser
>>>>>>>>>>>>> kernelization process
>>>>>>>>>>>>> https://inria.hal.science/hal-03404798/file/hipc2021.pdf
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>> And the x86 language says the same thing,
>>>>>>>>>>>>
>>>>>>>>>>>> YOU are just a liar, as proved by the fact that you can not 
>>>>>>>>>>>> give the Diagonalization proof you claimed you had.
>>>>>>>>>>>>
>>>>>>>>>>>> Sorry, you are just too stupid to understand.
>>>>>>>>>>>
>>>>>>>>>>> You continue to assume that you can simply disagree
>>>>>>>>>>> with the x86 language. My memory was refreshed that
>>>>>>>>>>> called you stupid would be a sin according to Christ.
>>>>>>>>>>> I really want to do the best I can to repent.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> But I am NOT disagreeing with the x86 language.
>>>>>>>>>>
>>>>>>>>>> Can you point out what fact of it I am disagreing about it?
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> You keep trying to get away with saying that the simulation is
>>>>>>>>> incorrect when the semantics of the x86 language conclusively
>>>>>>>>> proves that it is correct.
>>>>>>>>
>>>>>>>> Nope, and x86n emulation is only fully correct if it continues 
>>>>>>>> to the final end. 
>>>>>>>
>>>>>>> void Infinite_Loop()
>>>>>>> {
>>>>>>>    HERE: goto HERE;
>>>>>>> }
>>>>>>>
>>>>>>> Why do you say such ridiculously stupid things that you are are 
>>>>>>> false?
>>>>>>>
>>>>>>
>>>>>> Your Infinite_Loop does not apply. For a two cycle recursive 
>>>>>> simulation
>>>>>>
>>>>>
>>>>> This says nothing about two cycles nitwit.
>>>>>
>>>>> _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.
>>>>>
>>>>>
>>>>
>>>> You told us that HHH aborts its simulation after two cycles. 
>>>
>>> The above HHH refers to every pure function that correctly
>>> emulates one of more steps of DDD.
>>>
>>
>> As long as there are a finite number of steps, there is no infinite 
>> recursion. The problem remains that HHH cannot simulate itself correctly.
> 
> 
> void Infinite_Loop()
> {
>    HERE: goto HERE;
> }
> 
> Likewise infinite loops are not infinite loops thus
> contradicting the law of identity.
> 

Bad example, although I know you think two equals infinite.

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

You (and your simulator) will probably think Finite_Recursion is also an 
infinite recursion.