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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: Every D(D) simulated by H presents non-halting behavior to H ...
Date: Thu, 9 May 2024 10:24:40 +0300
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On 2024-05-08 19:16:15 +0000, olcott said:

> On 5/8/2024 11:11 AM, Mikko wrote:
>> On 2024-05-08 12:57:07 +0000, olcott said:
>> 
>>> On 5/8/2024 3:46 AM, Mikko wrote:
>>>> On 2024-05-07 15:40:32 +0000, olcott said:
>>>> 
>>>>> On 5/7/2024 6:18 AM, Richard Damon wrote:
>>>>>> On 5/7/24 3:30 AM, Mikko wrote:
>>>>>>> On 2024-05-06 18:28:37 +0000, olcott said:
>>>>>>> 
>>>>>>>> On 5/6/2024 11:19 AM, Mikko wrote:
>>>>>>>>> On 2024-05-05 17:02:25 +0000, olcott said:
>>>>>>>>> 
>>>>>>>>>> The x86utm operating system: https://github.com/plolcott/x86utm enables
>>>>>>>>>> one C function to execute another C function in debug step mode.
>>>>>>>>>> Simulating Termination analyzer H simulates the x86 machine code of its
>>>>>>>>>> input (using libx86emu) in debug step mode until it correctly matches a
>>>>>>>>>> correct non-halting behavior pattern proving that its input will never
>>>>>>>>>> stop running unless aborted.
>>>>>>>>>> 
>>>>>>>>>> Can D correctly simulated by H terminate normally?
>>>>>>>>>> 00 int H(ptr x, ptr x)  // ptr is pointer to int function
>>>>>>>>>> 01 int D(ptr x)
>>>>>>>>>> 02 {
>>>>>>>>>> 03   int Halt_Status = H(x, x);
>>>>>>>>>> 04   if (Halt_Status)
>>>>>>>>>> 05     HERE: goto HERE;
>>>>>>>>>> 06   return Halt_Status;
>>>>>>>>>> 07 }
>>>>>>>>>> 08
>>>>>>>>>> 09 int main()
>>>>>>>>>> 10 {
>>>>>>>>>> 11   H(D,D);
>>>>>>>>>> 12 }
>>>>>>>>>> 
>>>>>>>>>> *Execution Trace*
>>>>>>>>>> Line 11: main() invokes H(D,D);
>>>>>>>>>> 
>>>>>>>>>> *keeps repeating* (unless aborted)
>>>>>>>>>> Line 03: simulated D(D) invokes simulated H(D,D) that simulates D(D)
>>>>>>>>>> 
>>>>>>>>>> *Simulation invariant*
>>>>>>>>>> D correctly simulated by H cannot possibly reach past its own line 03.
>>>>>>>>>> 
>>>>>>>>>> The above execution trace proves that (for every H/D pair of the
>>>>>>>>>> infinite set of H/D pairs) each D(D) simulated by the H that this D(D)
>>>>>>>>>> calls cannot possibly reach past its own line 03.
>>>>>>>>> 
>>>>>>>>> When you say "every H/D pair" you should specify which set of pairs
>>>>>>>>> you are talking about. As you don't, your words don't mean anything.
>>>>>>>>> 
>>>>>>>> 
>>>>>>>> Every H/D pair in the universe where D(D) is simulated by the
>>>>>>>> same H(D,D) that D(D) calls. This involves 1 to ∞ steps of D
>>>>>>>> and also includes zero to ∞ recursive simulations where H
>>>>>>>> H simulates itself simulating D(D).
>>>>>>> 
>>>>>>> "In the universe" is not a set. In typical set theories like ZFC there
>>>>>>> is no universal set.
>>>>>> 
>>>>> 
>>>>> This template defines an infinite set of finite string H/D pairs where 
>>>>> each D(D) that is simulated by H(D,D) also calls this same H(D,D).
>>>> 
>>>> No, it does not. D does not know which H simulates it or even whther is
>>>> simulated at all. D calls the one whose name is H. The association of D
>>>> with the H of the same pair, if desired, must be done outside of D.
>>>> 
>>> 
>>> The template specifies that D(D) is calling the same H(D,D)
>>> that invokes it. All instances conform to the template.
>> 
>> No, it does not. The "H" in the template is a fixed "H" that does
>> not chage to "main" when
>> 
>>  int main(void) {
>>    D(D);
>>    return 0;
>>  }
>> 
>> is executed.
>> 
> 
> *I can get to that point only after you agree with this point*
> *I can get to that point only after you agree with this point*
> *I can get to that point only after you agree with this point*

DOesn't matter. Others are not that limited.

> 00 int H(ptr x, ptr x)  // ptr is pointer to int function
> 01 int D(ptr x)
> 02 {
> 03   int Halt_Status = H(x, x);
> 04   if (Halt_Status)
> 05     HERE: goto HERE;
> 06   return Halt_Status;
> 07 }
> 08
> 09 int main()
> 10 {
> 11   H(D,D);
> 12 }
> 
> Any H/D pair matching the above template where D(D) is simulated
> by the same H(D,D) that it calls cannot possibly reach past its own
> line 03. Simple software engineering verified fact.
> 
> We also must mutually agree that whenever any executed H(D,D)
> stops simulating its input that no H ever returns any value to any
> simulated D.

The meaning of "no H" is not suffficiently well specified to exclude
every H that returns some value to some simulated D.

-- 
Mikko