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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 20:15:32 -0400
Organization: i2pn2 (i2pn.org)
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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.

> 
> The above is STEP TWO  of my four step proof. STEP ONE can be
> discarded if it is easy enough for you and others to accept
> this STEP TWO. I already covered most of STEP THREE.
> 

And thus your proof is shown to be just wrong.

Rmember, when we are talking about Turing Machines (and Compuation 
theory in general) a "Correct Simulation" means a simulation that is 
never aborted, and thus WILL correctly recreate the behavior of the 
machine being simulated.

That means the correct simulation of a non-halting machine description, 
IS non-halting.

That is why using "correct simulation" as your method of halt deciding 
doesn't work.

You need to use partial simulation that actually PROVES what the Correct 
Simulation (that isn't actually done) would do.

I.e. your wording of "Correct Simulation by H" automattically means H 
must fail for non-halting inputs, as if H DOES do a correct simulation 
(as the criteria states) it must be non-halting.

You need your H to try to analyze the Correct Simulation by a UTM (which 
isn't H) would never halt, but that gets you right back to where you 
started.