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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 21:19:48 -0400
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On 5/26/24 9:03 PM, olcott wrote:
> 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


Not. what you said. You do like to change meaning with a "paraphrase" 
which is actually just a way to LIE.

Typical method of deliberate liars.

your clause about a finite sequence of steps was about reaching a final 
state, NOT about the actual simulation, which was just defined a 
"Correct" which means like a UTM which means it doesn't abort until it 
reaches the final state (which you admit it won't, so the simulator is 
stuck being non-halting, and thus not a decider).


> 
> 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
> 

It can be, but the isn't a UTM, BY DEFINITION.

Look up the definition of a Universal Turing Machine.

https://en.wikipedia.org/wiki/Universal_Turing_machine

In short, the Universal Turing Machine, when given the proper 
description of the Machine it emulate, computes the same result as the 
machine described.

This can not be done if the UTM aborts after a finite number of steps 
without reaching the final state.

And also, there is nothing in the definition that the UTM actually step 
through the same virtual sequence of states of the machine it is 
emulating. Yes, that is a natural way to do it, but isn't required, 
because everything in this area of computation theory is about input to 
output mapping, and NOT about how it was done, except for the 
requirement that it was done with a finite algorithm processing the data 
given to it.



You are just proving your ignorance of the subject.