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

On 6/29/24 12:09 PM, olcott wrote:

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.

Nope, we are not disagreeing with the semantics of the x86 language, we are disagreeing with your misunderstanding of how it works.

 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.

No the x86 language "knows" NOTHING about H0 being a x86 emulator. It is just a function that maybe happens to be a partial x86 emulator, but that is NOT a fundamental result of it being H0.

 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.

It is construed as non-halting BECAUSE it has been shown that your H0 *WILL* terminate its PARTIAL emulation of the code it is emulating and returning.

 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.

Right, so H0 is REQUIRED to return, and thus if the termination analyser knows that H0 is a termination analyzer it knows that the call to H0 MUST return, and thus DDD must be a terminating program.
An H0 that doesn't know this, and can't figure out that H0 will return, but just keeps emulating H0 emulating its input will just fail to meet its own requirement to return.

 <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>

Right, and the only definition Professor Sipser uses for "Correct Simulation" is a simulation that EXACTLY REPRODUCES the behavior of the directly executed program represented by the input. Your H doesn't do that, nor correctly predicts the behavior of such a simulation of the input (since that behavior is to halt) so it can never proper avail itself of the second paragraph, so does so erroneously getting the wrong answer.

 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.

Except that the "N Steps of DDD correctly emulated" is NOT the definition of the "behavior" of the input DDD.
"inputs" Do not have "behavoir", that is a property of a program, so the input only "represents" that program, in this case the program DDD.
Programs are also a COMPLETE specification of the steps the program will execute when run, so you below code is NOT a complete representation of the program DDD, so has include implicitly the definition of the SPECIFIC HHH that it has been pair with. One aspect of this implicit pairing is that this input inlcudes that EXACT HHH as part of its definition (and thus is NOT a "template" so when you argue about giving it to some other machine HHH, either you are giving that HHH a DIFFERENT input, or that other HHH is giving this DDD paired with the original HHH.
The Behavior of this input, given your claim that HHH will "correctly" determine it to be non-halting and return that answer, is thus the behavior of the program DDD, including the code of THAT HHH that will return to it and return, and thus your HHH is just wrong about the non-terminating behavior of the input.
Looking at your arguement, we need to look at what you are claiming to do, and fully define it. Are you trying to make DDD just a template (which doesn't by itself have behavior), or has DDD been bound to the decider you are claiming.
In the first case, we can't make any logical determination about THIS input from the behavior of other deciders looking at OTHER inputs.
In the second case, when the other decider simulates THIS input longer, it will see that it DOES halt (since THIS HHH returns) so your argument that "No HHH" can emulate thins

 _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
 

Since it is clear that you are making the same mistakes in that paper as you made here, they don't need repeating.
You are just showing you don't understand any of the basics of the field you are trying to talk about.
You have programs that are not programs and trying to redefine what "Halting" means, which just proves you are a liar,