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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Peter Olcott seems to consistently lie about this
Date: Tue, 6 Aug 2024 07:50:28 -0400
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On 8/6/24 7:35 AM, olcott wrote:
> On 8/6/2024 3:07 AM, Mikko wrote:
>> On 2024-08-05 12:45:11 +0000, olcott said:
>>
>>> On 8/5/2024 2:27 AM, Mikko wrote:
>>>> On 2024-08-04 12:33:20 +0000, olcott said:
>>>>
>>>>> On 8/4/2024 2:15 AM, Mikko wrote:
>>>>>> On 2024-08-03 13:48:12 +0000, olcott said:
>>>>>>
>>>>>>> On 8/3/2024 3:06 AM, Mikko wrote:
>>>>>>>> On 2024-08-02 02:09:38 +0000, olcott said:
>>>>>>>>
>>>>>>>>> *This algorithm is used by all the simulating termination 
>>>>>>>>> analyzers*
>>>>>>>>> <MIT Professor Sipser agreed to ONLY these verbatim words 
>>>>>>>>> 10/13/2022>
>>>>>>>>>      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
>>>>>>>>>
>>>>>>>>>      H can abort its simulation of D and correctly report that D
>>>>>>>>>      specifies a non-halting sequence of configurations.
>>>>>>>>> </MIT Professor Sipser agreed to ONLY these verbatim words 
>>>>>>>>> 10/13/2022>
>>>>>>>>>
>>>>>>>>> DDD is correctly emulated by HHH according to the x86
>>>>>>>>> language semantics of DDD and HHH including when DDD
>>>>>>>>> emulates itself emulating DDD
>>>>>>>>>
>>>>>>>>> *UNTIL*
>>>>>>>>>
>>>>>>>>> HHH correctly determines that never aborting this
>>>>>>>>> emulation would cause DDD and HHH to endlessly repeat.
>>>>>>>>
>>>>>>>> The determination is not correct. DDD is a halting computation, as
>>>>>>>> correctely determined by HHH1 or simly calling it from main. It is
>>>>>>>> not possible to correctly determine that ha haling computation is
>>>>>>>> non-halting, as is self-evdent from the meaning of the words.
>>>>>>>>
>>>>>>>
>>>>>>> [Who here is too stupid to know that DDD correctly simulated
>>>>>>>   by HHH cannot possibly reach its own return instruction?]
>>>>>>
>>>>>> Who here is too stupid to know that whether DDD can reach its
>>>>>> own return instruction depends on code not shown below?
>>>>>>
>>>>>
>>>>> void DDD()
>>>>> {
>>>>>    HHH(DDD);
>>>>>    return;
>>>>> }
>>>>>
>>>>> It is stipulated that HHH is an x86 emulator the emulates
>>>>> N instructions of DDD where N is 0 to infinity.
>>>>
>>>> That is not stipulated above. Anyway, that stipulation would not
>>>> alter the correctness of my answer.
>>>>
>>>
>>> typedef void (*ptr)();
>>> int HHH(ptr P);
>>>
>>> void DDD()
>>> {
>>>    HHH(DDD);
>>>    return;
>>> }
>>>
>>> int main()
>>> {
>>>    HHH(DDD);
>>> }
>>>
>>> In other words you do not know C well enough to comprehend
>>> that DDD correctly simulated by any HHH cannot possibly reach
>>> its own "return" instruction halt state.
>>
>> You are lying again.
>>
> 
> I am hypothesizing. If you do know C well enough to agree then
> simply agree. What I said is a tautology thus disagreement <is> error.
> 

Except that you hypotesize with fussy words, and thus break the procedure.

IF "Correct simulation" means, as normal, one that runs to completion, 
then the statement is true, but any HHH that does abort isn't given the 
input that was proven to not abort, as the HHH that DDD calls is part of it.

IF "Correct Simulation" means, as you claim, that it can be partial, 
then the statement is just incorrect, the partial simulation may not 
reach the return, but the program that was being simulated, continues 
after that aborting, and reaches the point, and thus the statment isn't 
true.

Part of your problem is you confuse the partial simulation of a program 
with the behavior of that program, which is defined as it COMPLETE 
behavior, and thus not shown by a partial simulation of it.

THis matches your confusion between truth and knowledge, and you total 
lack of knowledge about most of what you talk about.

Sorry, you are just proving yourself to be a pathological liar.