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

On 8/2/2024 4:55 AM, Mikko wrote:

On 2024-07-31 16:07:34 +0000, olcott said:
 

On 7/31/2024 2:42 AM, Mikko wrote:

On 2024-07-30 14:21:02 +0000, olcott said:
>

On 7/30/2024 1:52 AM, Mikko wrote:

On 2024-07-29 14:07:53 +0000, olcott said:
>

HHH(Infinite_Recursion) and HHH(DDD) show the same non-halting
behavior pattern in their derived execution traces of their
inputs.

>
Hard to believe as their behaviour is so different and you don't
say what pattern the see.

>
*Its all in the part that you erased*
>
*Infinite_Recursion correctly emulated by HHH*
*THREE lines repeat with no conditional branch instructions*
Begin Local Halt Decider Simulation   Execution Trace Stored at:113934
[0000215a][00113924][00113928] 55         push ebp      ; 1st line
[0000215b][00113924][00113928] 8bec       mov ebp,esp   ; 2nd line
[0000215d][00113920][00002162] e8f8ffffff call 0000215a ; 3rd line
[0000215a][0011391c][00113924] 55         push ebp      ; 1st line
[0000215b][0011391c][00113924] 8bec       mov ebp,esp   ; 2nd line
[0000215d][00113918][00002162] e8f8ffffff call 0000215a ; 3rd line
Local Halt Decider: Infinite Recursion Detected Simulation Stopped
>
*DDD correctly emulated by HHH*
*FOUR lines repeat with no conditional branch instructions*
Begin Local Halt Decider Simulation   Execution Trace Stored at:113895
[00002177][00113885][00113889] 55         push ebp      ; 1st line
[00002178][00113885][00113889] 8bec       mov ebp,esp   ; 2nd line
[0000217a][00113881][00002177] 6877210000 push 00002177 ; push DDD
[0000217f][0011387d][00002184] e853f4ffff call 000015d7 ; call HHH
[00002177][0015e2ad][0015e2b1] 55         push ebp      ; 1st line
[00002178][0015e2ad][0015e2b1] 8bec       mov ebp,esp   ; 2nd line
[0000217a][0015e2a9][00002177] 6877210000 push 00002177 ; push DDD
[0000217f][0015e2a5][00002184] e853f4ffff call 000015d7 ; call HHH
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

>
As that part does not show the anwer it seems best to assume that they
do not see the same pattern or the pattern is not a real non-halting
behaviour pattern. Of course, if a proof is ever presented, we may need
to reconsider, but it is very unlikely that any such proof will ever
be presented.
>

>
A proof is any sequence of steps such that the conclusion
is a necessary consequence of its basis.
>
Proving that DDD correctly emulated by HHH matches the
infinite recursion behavior pattern.
(a) The semantics of the x86 language.
(b) the design of HHH provided below.
(c) The definition of infinite recursion provided below.
>
*Infinite recursion behavior pattern*
An emulated sequence of instructions that has no conditional
branch instructions in this sequence is exactly repeated when
it calls the same function with the same parameters again.
As long as the called function can be determined to never
return this proves infinite recursion.

 You have not proven that there is no conditional instructions
in the repeated cycle.

void DDD()
{
   HHH(DDD);
   return;
}
In other words you are too damn stupid to see that
DDD correctly emulated by HHH cannot possibly reach
its own return instruction?
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer