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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: Any honest person that knows the x86 language can see... predict
 correctly
Date: Wed, 31 Jul 2024 10:56:38 +0200
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Op 30.jul.2024 om 23:13 schreef olcott:
> On 7/30/2024 4:07 PM, joes wrote:
>> Am Tue, 30 Jul 2024 15:05:54 -0500 schrieb olcott:
>>> On 7/30/2024 1:48 PM, Fred. Zwarts wrote:
>>>> Op 30.jul.2024 om 17:14 schreef olcott:
>>>>> On 7/30/2024 9:51 AM, Fred. Zwarts wrote:
>>>>>> Op 30.jul.2024 om 16:21 schreef olcott:
>>>>>>> On 7/30/2024 1:52 AM, Mikko wrote:
>>>>>>>> On 2024-07-29 14:07:53 +0000, olcott said:
>>>>>>>>
>>>>>>>>> HHH(Infinite_Recursion) and HHH(DDD) show the same non-halting
>>>>>>>>> behavior pattern in their derived execution traces of their
>>>>>>>>> inputs.
>>>>>>>> Hard to believe as their behaviour is so different and you don't
>>>>>>>> say what pattern the see.
>>>>>>>
>>>>>>> *Its all in the part that you erased*
>>
>>>>>> We all see the differences between these two.
>>>>>
>>>>> They both correctly predict behavior that must be aborted to prevent
>>>>> the infinite execution of the simulating halt decider.
>>>>>
>>>> Except that the prediction for the second one is wrong. The simulation
>>>> of an aborting and halting function, like HHH, does not need to be
>>>> aborted.
>>> I proved otherwise. When the abort code is commented out then it keeps
>>> repeating again and again, thus conclusively proving that is must be
>>> aborted or HHH never halts.
>> But the abort is not commented out in the running code!
>>
> 
> I modified the original code by commenting out
> the abort and it does endlessly repeat just like
> HHH correctly predicted.
> 

So you proved that it can only produce a correct prediction for another 
input: the HHH that does not halt.
But when it must predict the behaviour of HHH that aborts and halts, it 
fails.
You think that is correct, because you are still dreaming of that other 
HHH that does not halt. Dreams are no substitute for logic proofs.