Re: People are still trying to get away with disagreeing with the semantics of the x86 language

On 6/30/24 6:41 PM, olcott wrote:

On 6/30/2024 4:57 PM, Richard Damon wrote:

On 6/30/24 5:48 PM, olcott wrote:

On 6/30/2024 2:31 PM, Richard Damon wrote:

On 6/30/24 1:18 PM, olcott wrote:

On 6/30/2024 3:42 AM, Mikko wrote:

On 2024-06-29 16:09:19 +0000, olcott said:
>

People are still trying to get away with disagreeing with
the semantics of the x86 language. That is isomorphic to
trying to get away with disagreeing with arithmetic.
>
typedef void (*ptr)();
int H0(ptr P);
>
void Infinite_Loop()
{
   HERE: goto HERE;
}
>
void Infinite_Recursion()
{
   Infinite_Recursion();
}
>
void DDD()
{
   H0(DDD);
}
>
int main()
{
   H0(Infinite_Loop);
   H0(Infinite_Recursion);
   H0(DDD);
}
>
Every C programmer that knows what an x86 emulator is knows
that when H0 emulates the machine language of Infinite_Loop,
Infinite_Recursion, and DDD that it must abort these emulations
so that itself can terminate normally.
>
When this is construed as non-halting criteria then simulating
termination analyzer H0 is correct to reject these inputs as
non-halting by returning 0 to its caller.
>
Simulating termination analyzers must report on the behavior
that their finite string input specifies thus H0 must report
that DDD correctly emulated by H0 remains stuck in recursive
simulation.
>
<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>
>
People are trying to get away with disagreeing with the semantics
of the x86 language by disagreeing that
>
The call from DDD to HHH(DDD) when N steps of DDD are correctly
emulated by any pure function x86 emulator HHH cannot possibly
return.
>
_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]
>
>
*A 100% complete and total rewrite of the prior paper*
https://www.researchgate.net/publication/381636432_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_P

>
Nothing above is or points to any evdence about the alleged disagreement.
>

>
Of course not. I only said the actual truth.
>
Richard just said that he affirms that when DDD correctly
simulated by HHH calls HHH(DDD) that this call returns even
though the semantics of the x86 language disagrees.

>
What in the sematics of the x86 language, which INCLUDES that ever instruction WILL be followed by the next instruction, says that the HHH that is calld by DDD won't eventually return.
>
Since you assert that HHH(DDD) called by main returns, then by your requreement that HHH be a "pure function" ALL copies of it will do the same thing.
>
Yes, the EMULATION of HHH by HHH, but that can not be the "behavior of the input" as that "behavior" depends on more than just the input.
>

>
Therefore DDD correctly simulated by HHH DOES NOT HALT.
Thus HHH correctly reports that DDD DOES NOT HALT.
>

>
And then it doesn't correct emulate the input, and thus is a LIAR.
>

 You already know that you are the liar here and are
lying about not knowing this.
 _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]
 The call from DDD to HHH(DDD) when N steps of DDD are
correctly emulated by any pure function x86 emulator
HHH at machine address 0000217a cannot possibly return.
 

The problem is that the N steps emulated by HHH are not, and CAN NOT be the "behavior of the input", but only the complete emulation of the input.
The problem with just the N steps, is that NOTHING about the input specifies that number N, so it isn't part of the input, and the "behavior of the input" must be determined by just the input (and what it represents).
Thus, your "argument" is based on invalid definitions, showing your utter stupidity.