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From: olcott <polcott333@gmail.com>
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 20:03:04 -0500
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On 5/26/2024 7:44 PM, Richard Damon wrote:
> On 5/26/24 8:21 PM, olcott wrote:
>> On 5/26/2024 7:15 PM, Richard Damon wrote:
>>> On 5/26/24 7:45 PM, olcott wrote:
>>>> On 5/26/2024 6:07 PM, Richard Damon wrote:
>>>>> 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.
>>>>
>>>> You always seem to make sure respond to a different set of words
>>>> than the words I actually said. This could be an honest mistake.
>>>>
>>>> *I SAID A CORRECT SIMULATION OF A FINITE NUMBER OF STEPS*
>>>
>>> No you didn't, not the last time.
>>>
>>> You said (H^) H^) correctly simulated by embbeded_H cannot ..."
>>>
>>> If embedded_H does a "Correct Simulation", then BY DEFINITION, it 
>>> never aborts.
>>>
>>> That it doesn't reach a final state in a finte number of steps, and 
>>> thus, that "Correct Simulation" was non-halting.
>>>
>>> (and your earlier statement tried to assert behavior of THIS H^ based 
>>> on the behaviof or a DIFFERENT H^ built on a diffferent embedded_H 
>>> with differet behaivor which is just unsound logic, as the two 
>>> machines are essentially unrelated as far as this behavior)
>>>
>>>>
>>>> *WHEN I EXPLICITLY STATE A FINITE NUMBER OF STEPS THEN YOU ARE*
>>>> *FLAT OUT WRONG TO SIMPLY ASSUME AN INFINITE NUMBER OF STEPS*
>>>
>>> Nope, you said it didn't reach a final state in a finite number of 
>>> steps, i.e the simulation is shown to be non-halting.
>>
>> *If you need to, reread that many times*
>>  >>>> 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.
> 
> Right, so if you claim embedded_H is actually DOING a "Correct 
> Simulation", then BY the DEFINITION of COMPUTATION THEORY, that is an 
> non-aborted simulation.
> 

CORRECT SIMULATION OF A FINITE NUMBER OF STEPS
CORRECT SIMULATION OF A FINITE NUMBER OF STEPS
CORRECT SIMULATION OF A FINITE NUMBER OF STEPS

UNLESS YOU CAN PROVE THAT A UTM CANNOT POSSIBLY BE ADAPTED TO COUNT
THE NUMBER OF STEPS AND THEN STOP I AM CORRECT AND YOU ARE DISHONEST

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer