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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 14:16:45 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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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. You changed only the inner
HHH's, not the outermost one, thus breaking the recursive simulation.

>> 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).
> If your name is Charlie and your leg gets amputated you are still
> yourself, you don't get renamed to Bill.
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.

-- 
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.