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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory,sci.logic
Subject: =?UTF-8?Q?Re=3A_A_simulating_halt_decider_applied_to_the_The_Peter_?=
 =?UTF-8?Q?Linz_Turing_Machine_description_=E2=9F=A8=C4=A4=E2=9F=A9?=
Date: Tue, 28 May 2024 07:34:21 -0400
Organization: i2pn2 (i2pn.org)
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On 5/27/24 11:24 PM, olcott wrote:
> On 5/27/2024 7:17 PM, Richard Damon wrote:
>> On 5/27/24 8:08 PM, olcott wrote:
>>> On 5/27/2024 5:44 PM, Richard Damon wrote:
>>>> On 5/27/24 6:32 PM, olcott wrote:
>>>>> On 5/27/2024 4:21 PM, Richard Damon wrote:
>>>>>> On 5/27/24 3:45 PM, olcott wrote:
>>>>>>> On 5/27/2024 11:33 AM, Richard Damon wrote:
>>>>>>>> On 5/27/24 12:22 PM, olcott wrote:
>>>>>>>>> On 5/27/2024 10:58 AM, Richard Damon wrote:
>>>>>>>>>> On 5/27/24 11:46 AM, olcott wrote:
>>>>>>>>>>> On 5/27/2024 10:25 AM, Richard Damon wrote:
>>>>>>>>>>>> On 5/27/24 11:06 AM, olcott wrote:
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> typedef int (*ptr)();  // ptr is pointer to int function in C
>>>>>>>>>>> 00       int H(ptr p, ptr i);
>>>>>>>>>>> 01       int D(ptr p)
>>>>>>>>>>> 02       {
>>>>>>>>>>> 03         int Halt_Status = H(p, p);
>>>>>>>>>>> 04         if (Halt_Status)
>>>>>>>>>>> 05           HERE: goto HERE;
>>>>>>>>>>> 06         return Halt_Status;
>>>>>>>>>>> 07       }
>>>>>>>>>>> 08
>>>>>>>>>>> 09       int main()
>>>>>>>>>>> 10       {
>>>>>>>>>>> 11         H(D,D);
>>>>>>>>>>> 12         return 0;
>>>>>>>>>>> 13       }
>>>>>>>>>>>
>>>>>>>>>>> The above template refers to an infinite set of H/D pairs 
>>>>>>>>>>> where D is
>>>>>>>>>>> correctly simulated by either pure simulator H or pure 
>>>>>>>>>>> function H. This
>>>>>>>>>>> was done because many reviewers used the shell game ploy to 
>>>>>>>>>>> endlessly
>>>>>>>>>>> switch which H/D pair was being referred to.
>>>>>>>>>>>
>>>>>>>>>>> *Correct Simulation Defined*
>>>>>>>>>>>     This is provided because many reviewers had a different 
>>>>>>>>>>> notion of
>>>>>>>>>>>     correct simulation that diverges from this notion.
>>>>>>>>>>>
>>>>>>>>>>>     A simulator is an x86 emulator that correctly emulates 1 
>>>>>>>>>>> to N of the
>>>>>>>>>>>     x86 instructions of D in the order specified by the x86 
>>>>>>>>>>> instructions
>>>>>>>>>>>     of D. This may include M recursive emulations of H 
>>>>>>>>>>> emulating itself
>>>>>>>>>>>     emulating D.
>>>>>>>>>>
>>>>>>>>>> And how do you apply that to a TEMPLATE that doesn't define 
>>>>>>>>>> what a call H means (as it could be any of the infinite set of 
>>>>>>>>>> Hs that you can instantiate the template on)?
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> *Somehow we got off track of the subject of this thread*
>>>>>>>>
>>>>>>>> I note that YOU keep on switching between your C program and 
>>>>>>>> Turing Machines.
>>>>>>>>
>>>>>>>> Note, per the implications that you implicitly agreed to (by not 
>>>>>>>> even trying to refute) the two systems are NOT equivalents of 
>>>>>>>> each other.
>>>>>>>>
>>>>>>>
>>>>>>> (1) I think you are wrong. I have not seen any of your
>>>>>>> reasoning that was not anchored in false assumptions.
>>>>>>> Your make fake rebuttal is to change the subject.
>>>>>>>
>>>>>>> (2) It does not matter my proof is anchored in the Linz
>>>>>>> proof and the H/D pairs are only used to have a 100% concrete
>>>>>>> basis to perfectly anchor things such as the correct meaning
>>>>>>> of D correctly simulated by H so that people cannot get away
>>>>>>> with claiming that an incorrect simulation is correct.
>>>>>>>
>>>>>>> int main() { D(D); } IS NOT THE BEHAVIOR OF D CORRECTLY SIMULATED 
>>>>>>> BY H.
>>>>>>> One cannot simply ignore the pathological relationship between H 
>>>>>>> and D.
>>>>>>>
>>>>>>>>>
>>>>>>>>> When Ĥ is applied to ⟨Ĥ⟩
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>
>>>>>>>>>   Ĥ copies its own Turing machine description: ⟨Ĥ⟩
>>>>>>>>>   then invokes embedded_H that simulates ⟨Ĥ⟩ with ⟨Ĥ⟩ as input.
>>>>>>>>>
>>>>>>>>> For the purposes of the above analysis we hypothesize that
>>>>>>>>> embedded_H is either a UTM or a UTM that has been adapted
>>>>>>>>> to stop simulating after a finite number of steps of simulation.
>>>>>>>>
>>>>>>>> And what you do mean by that?
>>>>>>>>
>>>>>>>> Do you hypothesize that the original H was just a pure UTM,
>>>>>>>
>>>>>>> The original proof does not consider the notion of a simulating
>>>>>>> halt decider so I have to begin the proof at an earlier stage
>>>>>>> than any definition of H.
>>>>>>
>>>>>> The biggest problem is that the input to the Turing machine 
>>>>>> decider H is the description of a Turing Machine H^, which is a 
>>>>>> SPECIFIC machine, 
>>>>>
>>>>> When you say "specific machine" you don't mean anything like a
>>>>> 100% completely specified sequence of state transitions encoded
>>>>> as a single unique finite string.
>>>>
>>>> Mostly.
>>>>
>>>> There doesn't need to be a unique finite string, but it is a 100% 
>>>> completely specified state transition/tape operation table.
>>>>
>>>
>>> When Ĥ is applied to ⟨Ĥ⟩
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>
>>> In other words Linz did not prove that there are no set
>>> of state transitions specified by ⊢* that derives the
>>> correct halt status of ⟨Ĥ⟩ ⟨Ĥ⟩.
>>>
>>> He only said there there is one specific machine that
>>> gets the wrong answer.
>>>
>>
>> He STARTS with a proof that one specific (but arbitrary) machine gets 
>> the wrong answer.
>>
> 
> *Not exactly, you are misreading this*
> 
> The domain of this problem is to be taken as the set of all Turing
> machines and all w; that is, we are looking for a single Turing machine
> that, given the description of an arbitrary M and w, will predict
> whether or not the computation of M applied to w will halt

  *** a single Turing Machine ***

not singular

> 
> ...
> 
> Proof: We assume the contrary, namely that there exists an algorithm,
> and consequently some Turing machine H, that solves the halting problem
> https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

  *** some Turing Machine ***

Note singular

> 
> Ordinary existential quantification looks for at least one
> element not exactly one element:

But you can look for at least one by looking for one without an 
assumption that it is unique.

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