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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic,comp.ai.philosophy
Subject: Re: People are still trying to get away with disagreeing with the
 semantics of the x86 language
Date: Sat, 29 Jun 2024 13:47:56 -0500
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On 6/29/2024 1:38 PM, Richard Damon wrote:
> On 6/29/24 2:06 PM, olcott wrote:
>> On 6/29/2024 12:59 PM, Richard Damon wrote:
>>> On 6/29/24 1:17 PM, olcott wrote:
>>>> On 6/29/2024 11:45 AM, Richard Damon wrote:
>>>>> 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.
>>>>>
>>>>
>>>> *According to the professor Sipser approved criteria YES IT IS*
>>>>
>>>
>>> Nope. Where dp you see that in what he says? Remember, you need to 
>>> interpret the words by what he means them to say.
>>>
>>> His ONLY definition of "Correct Simulation" is a simulation that 
>>> exactly recreates the behavior of the program described by the input, 
>>> and that in one that does not stop its simulation. So, NOT your "N Step"
>>>
>>
>> *N steps of correct simulation are specified*
>> H correctly simulates its input D until H
>> H correctly simulates its input D until H
>> H correctly simulates its input D until H
>> H correctly simulates its input D until H
> 
> Which does not determine the ACTUAL behavor
> 

_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]

That you already know that it does prove that DDD correctly
emulated by HHH would never stop running unless aborted
or out-of-memory error

*proves that you are trying to get away with a bald-faced lie*
I really hope that you repent before it is too late.


-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer