Re: Any honest person that knows the x86 language can see... predict correctly

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.