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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory,sci.logic
Subject: Olcott can't tell the difference between a machine and an infinte set
 of machines
Date: Fri, 10 May 2024 17:59:51 -0400
Organization: i2pn2 (i2pn.org)
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On 5/10/24 5:43 PM, olcott wrote:
> On 5/10/2024 4:19 PM, Richard Damon wrote:
>> On 5/10/24 5:11 PM, olcott wrote:
>>> On 5/10/2024 3:50 PM, Richard Damon wrote:
>>>> On 5/10/24 4:27 PM, olcott wrote:
>>>>> On 5/10/2024 3:09 PM, Richard Damon wrote:
>>>>>> On 5/10/24 1:49 PM, olcott wrote:
>>>>>>> On 5/10/2024 11:12 AM, Richard Damon wrote:
>>>>>>>> On 5/10/24 11:50 AM, olcott wrote:
>>>>>>>>> On 5/10/2024 9:18 AM, Richard Damon wrote:
>>>>>>>>>> On 5/9/24 11:10 PM, olcott wrote:
>>>>>>>>>>> On 5/9/2024 9:31 PM, Richard Damon wrote:
>>>>>>>>>>>> On 5/9/24 11:38 AM, olcott wrote:
>>>>>>>>>>>>> On 5/8/2024 8:38 PM, immibis wrote:
>>>>>>>>>>>>>> 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!
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
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