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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: Every D(D) simulated by H presents non-halting behavior to H ###
Date: Wed, 8 May 2024 22:46:50 -0500
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On 5/8/2024 10:05 PM, Mike Terry wrote:
> On 08/05/2024 20:05, 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.
>>
> 
> I'll respond with my assessment on this, provided you agree in advance 
> that you won't quote me elsewhere [in other threads/forums] in support 
> of your claims.  Not that I can really enforce this, but I think for the 
> most part you are basically honest, and would try to keep an agreement 
> you made on this, if you chose to make one.
> 

That is really great Mike, you have been a wonderful help.
I will agree not to quote you anywhere else but these two
forums and I am nearly certain that I never quoted anyone
else from these forums anywhere else besides these forums.

> You understand the reason I ask this:  you are unfortunately completely 
> unable to judge what other people say to you, and as soon as you 
> (mis)interpret the smallest thing as supporting some part of your 
> argument you will (mis)quote "Mike Terry [or whoever] agrees that 
> [something I did not agree to, or some literal quote taken out of 
> context, which misrepresents my actual opinion]."
> 

OK, so I agree that I will only quote you using the message ID
of your reply with its time/date stamp and I will only quote
you in these forums. I am at least 99% certain that I never
named anyone in these forums in any other forums or any other
means of communication.

Your greatest help was to confirm my intuition that a simulating
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