Re: Hypothetical possibilities --- Complete Proof

Op 01.aug.2024 om 14:20 schreef olcott:

On 8/1/2024 3:10 AM, Fred. Zwarts wrote:

Op 31.jul.2024 om 23:23 schreef olcott:

On 7/31/2024 3:01 PM, Fred. Zwarts wrote:

Op 31.jul.2024 om 17:14 schreef olcott:

On 7/31/2024 3:44 AM, Fred. Zwarts wrote:

Op 31.jul.2024 om 06:09 schreef olcott:

>
  machine   stack     stack     machine    assembly
  address   address   data      code       language
  ========  ========  ========  =========  =============
[00002192][00103820][00000000] 55         push ebp
[00002193][00103820][00000000] 8bec       mov ebp,esp
[00002195][0010381c][00002172] 6872210000 push 00002172 ; push DDD
[0000219a][00103818][0000219f] e833f4ffff call 000015d2 ; call HHH(DDD)
New slave_stack at:1038c4
>
We don't show any of HHH and show the execution trace of
of just DDD assuming that HHH is an x86 emulator.

>
This assumption is incorrect if it means that HHH is an unconditional simulator that does not abort.

This algorithm is used by all the simulating termination analyzers:
<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
>
     H can abort its simulation of D and correctly report that D
     specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>

>
So, Sipser only agreed to a correct simulation, not with an incorrect simulation that violates the semantics of the x86 language by skipping the last few instructions of a halting program.
>

>
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
int main()
{
   HHH(DD);
}
>
DD correctly emulated by HHH cannot possibly reach its own
second line. I switched to DDD correctly emulated by HHH

>
But it has been proven that no such HHH exists that simulates itself correctly. So, talking about a correct simulation by HHH is vacuous word salad.
>

because only C experts understood the above example and we
never had any of those here.

>
There are many C experts that looked at it, but you only got critic, because you keep hiding important properties of HHH, which made the conclusion impossible.

 The following is all that is needed for 100% complete proof
that HHH did emulate DDD correctly according to the semantics
of the x86 language and did emulate itself emulating DDD
according to these same semantics.

You are repeating the same false claim with out any self-reflection. It has been pointed out that there are many errors in this proof.
Why repeating such errors?

 _DDD()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov ebp,esp   ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404     add esp,+04
[00002182] 5d         pop ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
 Begin Local Halt Decider Simulation   Execution Trace Stored at:1138cc
[00002172][001138bc][001138c0] 55         push ebp      ; housekeeping
[00002173][001138bc][001138c0] 8bec       mov ebp,esp   ; housekeeping
[00002175][001138b8][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][001138b4][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)

The trace stops and hides what happens when 000015d2 is called.
Olcott is hiding the conditional branch instructions in the recursion.

New slave_stack at:14e2ec
[00002172][0015e2e4][0015e2e8] 55         push ebp      ; housekeeping
[00002173][0015e2e4][0015e2e8] 8bec       mov ebp,esp   ; housekeeping
[00002175][0015e2e0][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][0015e2dc][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)

Again the trace stops and hides what happens when 000015d2 is called.
Olcott is hiding the conditional branch instructions in the recursion.

Local Halt Decider: Infinite Recursion Detected Simulation Stopped

The program prints an erroneous message about an infinite recursion, when only two recursions are seen. It is clear the that the simulated HHH would need only one cycle more to return to DDD.
No other conclusion is possible than that HHH performs an incomplete and, therefore, incomplete simulation.

 Now that we have established that HHH did emulate
DDD correctly the above is also 100% complete proof
that HHH must abort its correct emulation of DDD or
neither HHH nor DDD will ever stop running.

'Must abort ..' is no proof that aborting will results in a correct simulation. At most it proves that not-aborting is incorrect, as well.
No matter how much olcott wants it to be correct, or how many times olcott repeats that it is correct, it does not change the fact that such a simulation is incorrect, because it is unable to reach the end of a halting program.
Olcott's own claim that the simulated HHH does not reach its end confirms it. The trace he has shown also proves that HHH cannot reach the end of its own simulation. So, his own claims prove that it is true that HHH cannot possibly simulate itself up to the end, which makes the simulation incomplete and, therefore, incorrect.
Dreams are no substitute for logic proofs.
Sipser would agree that this incorrect simulation cannot be used to detect a non-halting behaviour.

 This has always been complete proof for anyone that knows
the semantics of the x86 language sufficiently well.
 

No, anyone with sufficient knowledge of the semantics of the x86 language, sees that the simulation is incomplete and therefore incorrect. It is clear who lacks knowledge of the semantics of the x86 language.
The sad thing is that you don't want to learn what these semantics mean.