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

Op 31.jul.2024 om 23:27 schreef 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.
>

>
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.
 

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.

But then you also changed the input.
A better proof is the simulation of HHH by HHH1, which shows that no abort is needed to simulate HHH.

When we don't remove the abort code it gets aborted.
An average first year comp sci student can get this.
 

We are discussion the input given to HHH of a HHH that aborts.
You are changing the subject to another input
A better idea would be to change
   if (Root)
to
   if (!Root)
in the code of HHH
that would come close to a simulator that does not abort which simulates a HHH that does abort. That one will show you that no abort is needed.