Re: Disagreeing with tautologies is always incorrect

On 5/30/25 10:59 AM, olcott wrote:

On 5/30/2025 3:36 AM, Fred. Zwarts wrote:

Op 29.mei.2025 om 18:00 schreef olcott:

On 5/29/2025 10:46 AM, Fred. Zwarts wrote:

Op 29.mei.2025 om 17:37 schreef olcott:

HHH is a simulating termination analyzer that uses
an x86 emulator to emulate its input. HHH is capable
of emulating itself emulating DDD.
>
HHH is executed within the x86utm operating system
that enables any C function to execute another C
function in debug step mode.
>
*Here is the fully operational code*
https://github.com/plolcott/x86utm/blob/master/Halt7.c
>
void DDD()
{
   HHH(DDD);
   return;
}
>
_DDD()
[00002192] 55             push ebp
[00002193] 8bec           mov ebp,esp
[00002195] 6892210000     push 00002192
[0000219a] e833f4ffff     call 000015d2  // call HHH
[0000219f] 83c404         add esp,+04
[000021a2] 5d             pop ebp
[000021a3] c3             ret
Size in bytes:(0018) [000021a3]
>
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If simulating halt decider H correctly simulates its
     input D until H correctly determines that its simulated D
     would never stop running unless aborted then
>
It is a tautology that any input D to termination
analyzer H that *would never stop running unless aborted*
DOES SPECIFY NON-TERMINATING BEHAVIOR.
>

Indeed, but the input given to HHH is a pointer to memory starting with the code of DDD. DDD has the addresses of the functions it calls, so those functions are also available to HHH.

>
None-the-less the tautology remains true and thus
cannot be correctly denied.
>

This program includes the code where the program aborts and halts. If HHH would correctly analyse this input, it would see that no abort is needed.

>
If the outermost HHH does not abort its simulated DDD
then the simulated DDD, the executed DDD() and the executed
HHH() never stop running. That you cannot understand that
this is factual is no rebuttal what-so-ever.
>

But the innermost HHH has code to abort. Whether that is good or bad is not interesting, the point is: it is there.

 You are too dumb to know that this code is unreachable.
Either the outermost HHH aborts or none of them abort
because they all of the exact same code.
 

Yes, the code is unreachable because the call instruction is unsimulatable.
Fix that by including the code of the HHH that is claimed to give the right answer, and then the code is reachable, just not by the partial simulation done by that HHH.
Thus, you just prove that you "logic" is built on error and lies.

So, there is no need for the outermost HHH to abort and by this bug it misses the fact the innermost HHH already has an abort in its code.
That you cannot understand the verified fact that the simulation of an aborting simulator does not need an abort. Remove the non-aborting HHH from your brain. It confuses your thinking. There is no such HHH in this problem. There is only a buggy HHH that aborts, which therefore does not need an abort when simulated.