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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: HHH maps its input to the behavior specified by it --- never
 reaches its halt state ---
Date: Fri, 9 Aug 2024 11:19:03 -0500
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On 8/9/2024 10:27 AM, Fred. Zwarts wrote:
> Op 09.aug.2024 om 15:41 schreef olcott:
>> On 8/9/2024 1:39 AM, Fred. Zwarts wrote:
>>> Op 09.aug.2024 om 05:03 schreef olcott:
>>>> On 8/8/2024 9:52 PM, Richard Damon wrote:
>>>>> On 8/8/24 9:15 AM, olcott wrote:
>>>>>>
>>>>>> void DDD()
>>>>>> {
>>>>>>    HHH(DDD);
>>>>>>    return;
>>>>>> }
>>>>>>
>>>>>> Each HHH of every HHH that can possibly exist definitely
>>>>>> *emulates zero to infinity instructions correctly* In
>>>>>> none of these cases does the emulated DDD ever reach
>>>>>> its "return" instruction halt state.
>>>>>>
>>>>>> *There are no double-talk weasel words around this*
>>>>>> *There are no double-talk weasel words around this*
>>>>>> *There are no double-talk weasel words around this*
>>>>>>
>>>>>> There is no need to show any execution trace at the x86 level
>>>>>> every expert in the C language sees that the emulated DDD
>>>>>> cannot possibly reaches its "return" instruction halt state.
>>>>>>
>>>>>> Every rebuttal that anyone can possibly make is necessarily
>>>>>> erroneous because the first paragraph is a tautology.
>>>>>>
>>>>>>
>>>>>
>>>>> Nope, it is a lie based on comfusing the behavior of DDD which is 
>>>>> what "Halting" is.
>>>>>
>>>>
>>>> Finally something besides
>>>> the strawman deception,
>>>> disagreeing with a tautology, or
>>>> pure ad hominem.
>>>>
>>>> You must first agree with everything that I said above
>>>> before we can get to this last and final point that it
>>>> not actually directly referenced above.
>>>>
>>>> *Two key facts*
>>>> (a) The "return" instruction is the halt state of DDD.
>>>> (b) DDD correctly emulated by any HHH never reaches this state.
>>> There is no correct simulation of HHH by itself. HHH cannot possibly 
>>> simulate itself correctly. A correct simulation of a halting program 
>>> must reach this state.
>>
>> Try and show how it is incorrect.
> We proved it many times, but it seems you do not understand it. It is 
> unclear to me whether you are unable to understand it or unwilling to 
> understand it. But I will go another mile.
> 
> (HHH is required to halt, so we don't need to consider the non-aborting 
> HHH.)
> 1) HHH is programmed to abort after N cycles. When HHH simulates itself, 
> the simulating HHH aborts after N cycles.

That is false. AKA counter-factual.

> 2) Note, we are looking at the HHH that aborts after N cycles, not the 
> one that does not abort. Both the simulating and the simulated HHH use 
> the same algorithm, namely, the algorithm to abort after N cycles.

When we look at every HHH that can possibly exist then we
see that DDD correctly emulated by each one of these cannot
possibly reach its own "return" instruction halt state.

That you can't understand this seems to indicate a lack
of software engineering skills.

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer