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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: Mon, 27 May 2024 12:33:28 -0400
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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.

> 
> 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, in which 
case we have previously shown that H (H^) (H^) will never answer, and 
thus isn't a correct decider. This is the meaning if you change the 
contents of the description (H^) to use the new hypothisized embedded_H.

Or are you hypothesizing that we give the input to the embedded_H that 
is part of H^ the same input as the actual embedded_H, but then we see 
that this UTM will reach the final state, as since H (H^) (H^) has been 
said to go to qn, and thus the embedded_H that was actually in H^, we 
see that H^ (H^) will go to H^.qn and the simulation will halt.

> 
>  From this we can see that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H
> cannot possibly reach its own simulated final state of ⟨Ĥ.qn⟩ and halt
> in an infinite or any finite sequence of correctly simulated steps.

Which means you are doing the first, and thus it was looking at a 
DIFFERENT H^ then the one given to the actual H and embedded_H, and thus 
its answer is not applicable.

You are trying to describe the properties of a cat by studing a 10-story 
office building.

The fact that you claim that no embedded_H can reach the final state in 
its PARTIAL simulation doesn't prove anything, and in fact, when we do 
the SECOND version of the hypothetical, we see that a different 
simulator, given the same input, CAN reach the final state, and thus the 
input represents a Halting machine.

> 
> When an infinite number of steps is not enough then we can definitely
> conclude that less than an infinite numbers of steps is also not enough.
> 

But you use invalid logic that conflates the behavior of different machines.

The only machine you actually show to be non-halting, is the one built 
on an H that is actually just a plain UTM, and never aborts, and that H 
fails to answer.

All other versions of H^ can be shown to halt, as the H they are built 
on was described as going to H.qn, and thus H^ goes to H^.qn and halts.

You are still working with desceptively vague definitions that you try 
to work as having two different meanings in different parts of the 
proof, which is just invalid logic, showing you just don't understand 
the nature of logic, or the field you are talking about.