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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: Mike Terry Proves --- How the requirements that Professor Sipser
 agreed to are exactly met
Date: Tue, 20 May 2025 19:46:19 +0200
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Op 20.mei.2025 om 16:37 schreef olcott:
> On 5/20/2025 2:06 AM, Mikko wrote:
>> On 2025-05-20 04:20:54 +0000, olcott said:
>>
>>> <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
>>>
>>> Do you understand that we are only evaluating whether
>>> or not HHH/DDD meets this above criteria?
>>
>> I do understand that the meaning of the behaviour is not mentioned
>> in the creteria and is therefore irrelevant, an obvious consequence
>> of which is that your "WRONG!" above is false.
>>
> 
> *H correctly simulates its input D until*
> specifies that HHH must simulate DDD according
> to the meaning of the rules of the x86 language.
> 
> The meaning of every step of the behavior is
> precisely specified by the x86 language.
> 
> _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]
> 
> *H correctly simulates its input D*
> 00002172 00002173 00002175 0000217a
> H correctly simulates itself simulating DDD
> 00002172 00002173 00002175 0000217a
> 
> *until H correctly determines that its simulated D*
> *would never stop running unless aborted*

That is a wild guess of HHH, not a correct determination. When it sees 
the call to HHH and we know that HHH halts, we know that there is only a 
finite recursion, so the 'would never stop running' exists only in your 
dreams.
The input is a finite string that includes the code of Halt7.c, which 
specifies that the simulation will abort. So, HHH is wrong when it 
assumes that an abort is needed for this input to prevent  a never stop 
running.
Face the facts, not your dreams. Try a real argument, instead a 
repetition of your dream. Try to get out of rebuttal mode.

> 
> H sees DDD call the same function with the same
> parameter and there are no conditional branch
> instructions from the beginning of DDD to calling
> HHH(DDD) again. This repeating pattern proves
> non-termination.
> 
> 

HHH does not even see a full cycle, so it cannot know that there are no 
conditional branches in the cycle. You can view a full cycle in 
different ways:
1) from the first start of DDD up to the second start of DDD. The second 
beginning of DDD is reached after many steps of the simulation, which 
contains a lot of conditional branching instruction.
2) From the first start of HHH up to the second start of HHH. In this 
cycle there are also many conditional branch instructions within HHH.
So, it is misleading to say that there are no conditional branch 
instruction in the full cycle.
That a small part of the cycle does not have conditional branch 
instructions does not prove anything.
Face the facts. Stop repeating your dreams. Come out of rebuttal mode 
and try a serious honest dialogue.