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From: immibis <news@immibis.com>
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Subject: Re: Every D(D) simulated by H presents non-halting behavior to H ###
Date: Thu, 9 May 2024 03:38:38 +0200
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On 8/05/24 21:05, olcott wrote:
> On 5/8/2024 10:13 AM, Mike Terry wrote:
>> On 08/05/2024 14:01, olcott wrote:
>>> On 5/8/2024 3:59 AM, Mikko wrote:
>>>> On 2024-05-07 19:05:54 +0000, olcott said:
>>>>
>>>>> On 5/7/2024 1:54 PM, Fred. Zwarts wrote:
>>>>>> Op 07.mei.2024 om 17:40 schreef olcott:
>>>>>>> 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).
>>>>>>>
>>>>>>> These H/D pairs can be enumerated by the one to ∞ simulated steps 
>>>>>>> of D and involve zero to ∞ recursive simulations of H simulating 
>>>>>>> itself simulating D(D). Every time Lines 1,2,3 are simulated 
>>>>>>> again defines
>>>>>>> one more level of recursive simulation.
>>>>>>>
>>>>>>> 1st element of H/D pairs 1 step  of D  is simulated by H
>>>>>>> 2nd element of H/D pairs 2 steps of D are simulated by H
>>>>>>> 3rd element of H/D pairs 3 steps of D are simulated by H
>>>>>>>
>>>>>>> 4th element of H/D pairs 4 steps of D are simulated by H
>>>>>>> this begins the first recursive simulation at line 01
>>>>>>>
>>>>>>> 5th element of H/D pairs 5 steps of D are simulated by
>>>>>>> next step of the first recursive simulation at line 02
>>>>>>>
>>>>>>> 6th element of H/D pairs 6 steps of D are simulated by
>>>>>>> last step of the first recursive simulation at line 03
>>>>>>>
>>>>>>> 7th element of H/D pairs 7 steps of D are simulated by H
>>>>>>> this begins the second recursive simulation at line 01
>>>>>>
>>>>>> Is this the definition of the infinite set of H? We can think of 
>>>>>> many more simulations that only these.
>>>>>
>>>>> 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-one can possibly show one element of this set where D(D) reaches
>>>>> past its own line 03.
>>>>
>>>> If H is a decider of any kind then the D build from it reaches its line
>>>> 4 as numberd above. Whether the simulation of D by H reaches that line
>>>> is another question.
>>>>
>>>
>>> *My fully operational code proves otherwise*
>>>
>>> I seems like you guys don't have a clue about how infinite
>>> recursion works. You can run the code and see that I am correct.
>>>
>>> I have one concrete instance as fully operational code.
>>> https://github.com/plolcott/x86utm/blob/master/Halt7.c
>>> line 555 u32 HH(ptr P, ptr I) its input in on
>>> line 932 int DD(int (*x)())
>>
>> HH is completely broken - it uses a global variable which is allows HH 
>> to detect whether it is the outer HH or a nested (simulated) HH.  As a 
>> result, the nested HH behaves completely differently to the outer HH - 
>> I mean /completely/ differently: it goes through a totally separate "I 
>> am called in nested mode" code path!
>>
> 
> The encoding of HH is not the pure function that it needs to be to
> be a computable function.
> 
> *Maybe you can settle this*
> 
> The disagreement is entirely over an enormously much simpler thing.
> The disagreement is that Richard says that a D simulated by H could
> reach past its own line 03 and halt.

Here's the proof:

1. A simulation always produces an identical execution trace to the 
direct execution.

2. The direct execution reaches past line 03 and halts.

3. Therefore, the simulation reaches past line 03 and halts.

So far, you have only shown things that are not simulations, but you 
called them "simulations" anyway because you are a liar.