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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: Any honest person that knows the x86 language can see... predict
 correctly
Date: Wed, 31 Jul 2024 16:17:10 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Wed, 31 Jul 2024 10:02:26 -0500 schrieb olcott:
> On 7/31/2024 9:16 AM, joes wrote:
>> Am Wed, 31 Jul 2024 05:52:54 -0500 schrieb olcott:
>>> On 7/31/2024 3:54 AM, joes wrote:
>>>> Am Tue, 30 Jul 2024 16:13:55 -0500 schrieb 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:
>>>>
>>>>>>> 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.
>>>
>>>> Yes, and that modification makes HHH not call itself
>>> Not at all. It makes HHH stop aborting DDD.
>>> So that HHH and DDD endlessly repeat.
> 
>> Commenting out a section changes the program.
> This conclusively proving that this section was required.
When you put in the abort, it also appears in the simulated HHH.

>> You changed only the inner HHH's, not the outermost one, thus breaking
>> the recursive simulation.
> Not at all. I simply disabled the abort and this resulted in unlimited
> repetition non-halting behavior.
You did NOT change all calls to HHH.

>>>> but a different program. You'd need to also comment out the outermost
>>>> abort; then it wouldn't halt, but if you change HHH to abort, you
>>>> change all copies of it at the same time (to keep the recursive call
>>>> structure).
>> A program's identity changes with its code. It doesn't matter what I
>> label it in the source. I can define different functions with the same
>> name.
> To prove that a section of code is required we remove that section and
> the resulting endless repetition proves that the abort section was
> required to prevent the endless repetition.
Enough said.

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.