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

Am Wed, 31 Jul 2024 16:27:58 -0500 schrieb olcott:

On 7/31/2024 3:09 PM, Fred. Zwarts wrote:

Op 31.jul.2024 om 18:32 schreef olcott:

On 7/31/2024 11:17 AM, joes wrote:

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.
>

Yet this is unreachable in the same way that in a single file foot
race with everyone going the same speed and everyone 15 feet ahead of
the next person that the first person must win.

 
Yet that is no reason for the person in front to kill all other people,
because otherwise they would not stop running.
The first person will stop at the finish, the second person will stop
at the finish, the third .... etc.
 
There is no reason to assume that there are persons that will keep
running indefinitely.

They will run unto the finish.

The outermost HHH sees that it must abort one whole execution trace
sooner than the next inner HHH.

But it is wrong to assume that the simulated HHH would not have halted
when not aborted.

It has never been an assumption is has always been a tautology that has
always been over your head. Joes may be catching up with the Linz proof.

If one of them aborts, all of them do.

This is proved when HHH is simulated by a non-aborting simulator, such
as HHH1. A correct simulation shows that the simulated HHH does not
need to be aborted.

When we remove the abort code it keeps repeating. When we don't remove
the abort code it gets aborted.

Also, the simulated HHH aborts and the simulating HHH doesn't trigger.

--
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.