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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: Sun, 18 May 2025 15:30:17 -0400
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On 5/18/25 3:18 PM, olcott wrote:
> On 5/18/2025 2:08 PM, joes wrote:
>> Am Sun, 18 May 2025 12:28:05 -0500 schrieb olcott:
>>> On 5/18/2025 10:21 AM, Mike Terry wrote:
>>>> On 18/05/2025 10:09, Mikko wrote:
>>>>> On 2025-05-17 17:15:14 +0000, olcott said:
>>
>>>>>> HHH(DDD) does not base its decision on the actual behavior of DDD
>>>>>> after it has aborted its simulation of DDD, instead it bases its
>>>>>> decision on a different HHH/DDD pair that never aborts.
>>>>>
>>>>> This is why HHH does not satisfy "H correctly determines that its
>>>>> simulated D would never stop running unless aborted". If HHH bases its
>>>>> decision on anything else than what its actual input actually
>>>>> specifies it does not decide correctly.
>>>>>
>>>> Right.  It seems to be a recent innovation in PO's wording that he has
>>>> started using the phrase "..bases its decision on a different *HHH/DDD
>>>> pair* ..".
>>>>
>>> Thus SHD must report on a different SHD/Infinite_Loop pair where this
>>> hypothetical instance of itself never aborts.
>> This, the simulator. The input still calls the same real aborting HHH.
>>
>>> If H always reports on the behavior of its simulated input after it
>>> aborts then every input including infinite_loop would be determined to
>>> be halting.
>> Yes, that is why H is wrong.
>>
>>> Instead H must report on the hypothetical H/D input pair where the very
>>> same H has been made to not abort its input.
>> Just no.
>>
>>> *H correctly determines that its simulated D*
>>> *would never stop running unless aborted*
>>> by a hypothetical instance of itself that never aborts.
>> H does stop running when simulated without aborting, because it aborts.
>>
> 
> H is required to report on the behavior of D in the
> case where a hypothetical instance of itself never
> aborts its simulated D.

No, it is required to report on the behavior of D as it actually is, 
which means it must be a program (which yours isn't)

When you complere D per the proof you are trying to refute, it will be 
built on the H that you finally use to say gives the correct answer, 
which is the actual H that aborts and returns, and thus the only correct 
simulation of this input shows halting.

> 
> When the hypothetical H never aborts its simulated D then:
> (a) Simulated D  NEVER HALTS
> (b) Executed D() NEVER HALTS
> (c) Executed H() NEVER HALTS
> (d) Everything that H calls NEVER HALTS
> 

Only because you look at the wrong simuated D, because you D is just wrong.

You are just proving how ignorant you are of what you are talking about, 
and that you don't mind just stating out and out lies, because you are 
nothing but a pathological liar that doesn't care about what is actually 
true.