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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Mike Terry Proves --- How the requirements that Professor Sipser
 agreed to are exactly met
Date: Wed, 21 May 2025 07:06:49 -0400
Organization: i2pn2 (i2pn.org)
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On 5/20/25 11:05 PM, olcott wrote:
> On 5/20/2025 9:19 PM, Richard Damon wrote:
>> On 5/20/25 10:08 PM, olcott wrote:
>>> On 5/20/2025 8:06 PM, Mike Terry wrote:
>>>> On 20/05/2025 18:46, Fred. Zwarts wrote:
>>>>> 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.
>>>>>
>>>>
>>>> Yes, that all correct.  There are loads of conditional branch 
>>>> instructions performed by HHH as part of DDD.  This makes a nonsense 
>>>> of the implementation of PO's "infinite recursion" test.
>>>>
>>>> But there is a worse nonsense here:  even if there were indeed no 
>>>> conditional branches between the matching call statements in the 
>>>> simulation, THAT STILL WOULD NOT BE ENOUGH TO GUARANTEE INFINITE 
>>>> RECURSION!
>>>>
>>>
>>> You are assuming details of HHH that are not included
>>> in its specification. A DDD that is only simulated by
>>> HHH *is* infinite recursion.
>>
>>
>> But such an HHH isn't the needed decider, so not the DDD that we are 
>> looking at when we have a decider HHH.
>>
>> Sorry,
>>>
>>> void DDD()
>>> {
>>>    HHH(DDD);
>>>    return;
>>> }
>>>
>>> Any moron can see that DDD simulated by HHH cannot possibly halt.
>>> It does not matter how many steps of DDD are simulated by HHH.
>>> No DDD every reaches its own "return" statement final halt state.
>>
>> And any HHH that just simulates its input isn't a decider, and thus 
>> fails.
>>
> 
> I have to go one step at a time or people
> get completely overwhelmed.

But each step needs to be true and based on the rules.

> 
> So far everyone here including you right now made
> sure to dodge the above point, thus lack the mandatory
> prerequisites for moving on the to the next point.

No, it has been proved wrong many ways,

> 
> When an HHH emulates N steps of DDD,
> (no matter what the value of N is)
> DDD never halts and has the exact
> same behavior as HHH aborting DDD
> after N steps.
>

Except the DDD that you have defined can NOT be simulated past 4 steps, 
so I guess you think we can't count past 4.

And "Halting" is a property of PROGRAMS, of which you have admitted that 
DDD isn't one.

That is like saying HHH has proven that DDD isn't Green, since it 
doesn't have color either.