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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: Sun, 26 May 2024 19:07:39 -0400
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On 5/26/24 6:47 PM, olcott wrote:
> On 5/26/2024 3:20 PM, Richard Damon wrote:
>> On 5/26/24 3:14 PM, olcott wrote:
>>> When Ĥ is applied to ⟨Ĥ⟩
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>
>>> When we see that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H in an
>>> infinite number of steps cannot possibly reach its own simulated
>>> final state of ⟨Ĥ.qn⟩ and halt then we correctly deduce that the
>>> same thing applies when simulating halt decider embedded_H correctly
>>> simulates less than an infinite number of steps of ⟨Ĥ⟩ ⟨Ĥ⟩.
>>>
>>
>> Nope.
>>
>> Since we are talking about Turing Machines, your stipulated POOP 
>> definitions go away, 
> 
> https://www.liarparadox.org/Linz_Proof.pdf
> *Simplified the notation for Ĥ on the top of page three*
> and put back in the qy state shown in figure 12.2
> 
> 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.
> 
> It is an easily verified fact that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by 
> embedded_H cannot possibly reach its own simulated final state of
> ⟨Ĥ.qn⟩ in any finite sequence of steps.

Nope, since we are in Turing Machines, the term "Correctly Simulated" 
means, and can ONLY mean, the resuts of a UTM simulation, which BY 
DEFINITION is nopt aborted.

Thus, if you claim embedded_H actually does a correct (and thus 
UNABORTED) simulation, (and thus H does) then you H just fails to answer.

Thus, either you admit that your H/embedded_H fails to answer, or that 
you H/embedded_H don't actually do a correct simulation, so the term "a 
correct simulation by H" is just an impossible statement (something you 
seem to like to do).

We can fix it by dropping the "by H", but then the statement just isn't 
true, as if H applied to (H^) (H^) goes to qn (and thus also embedded_H) 
we see that UTM (H^) (H^) will go to H^.qn by the rules of construction 
of H^ and thus the "correct simulation" of the input will halt.

> 
> *If you want to lie about this or fail to understand this*
> *irrefutable truth that only makes yourself look foolish*

I am not lying, I an just apply the definition that you failed to learn. 
Your quoting of errors by rote about things you never learned about 
shows your ignorance of the topic.

> 
> Any academicians reading this post will further conclude
> that your insults and ad hominem attacks are quite pathetic.
> 

Nope, they will see the errors in your logic, due to the use of 
incorrect definitions.


Note, all these FACTS come out of the implications of your stipulated 
definitions of your termination analyzer, which you implicitily accepted 
by not even trying to rebute.

If you want to try to rebute them, show some actual reference from a 
scholerly accepted source that disagrees with what I am saying.