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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Any honest person that knows the x86 language can see... predict
 correctly
Date: Tue, 30 Jul 2024 21:21:44 -0400
Organization: i2pn2 (i2pn.org)
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On 7/30/24 5:13 PM, olcott wrote:
> 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.

But then it is a different input, as the code for the HHH that it calls 
MUST be considered as part of the input.

So, you are just proving yourself to be a stupid liar.

> 
>>>> This is proved when it is simulate by HHH1. HHH aborts after two
>>>> recursions, which is not an infinite execution.
>>>> We know you really, really wants it to be correct. So, you are cheating
>>>> by suppressing part of the trace, in order to hide the conditional
>>>> branch instructions in the second case. But no matter how much olcott
>>>> wants it to be correct, or how many times olcott repeats that it is
>>>> correct, it does not change the fact that such a simulation is
>>>> incorrect
>>
>