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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 23:15:52 -0400
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On 5/26/24 10:53 PM, olcott wrote:
> On 5/26/2024 9:30 PM, Richard Damon wrote:
>> On 5/26/24 10:06 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
>>>
>>>   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 such as Googolplex ^ Googolplex
>>> number of steps. https://en.wikipedia.org/wiki/Googolplex
>>
>> So, it can verify that if H / embedded_H was programmed not to halt it 
>> simulation, then the H^ built on that H will be non-halting, but that 
>> doesn't say anything about the DIFFERENT H^ built on an H that does 
>> abort its simulation and returns 0.
>>
> 
> You really do have an actual problem paying attention.
> It is not a mere trollish ruse. I poured engine oil
> into my power steering fluid Friday night so I can relate.

Shows your lack of paying attention to details, as you argument also shows.

> 
> As far as paying attention to words that have been written
> I read and reread many times until I am sure that I get it.
> Even then I get it wrong once in a while.

You mean most of time, Rereading doesn't help when you don't know the 
meaning of the words. Thats the problem with talking about a field you 
haven't studied.

> 
> When embedded_H correctly simulates a Googolplex ^ Googolplex steps
> of ⟨Ĥ⟩ ⟨Ĥ⟩ then embedded_H halts.

And thus embedded_H is NOT a UTM, and its simulation say NOTHING about 
the behavior of the machine it is simulating.

> 
> ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H cannot possibly reach its
> own simulated final state of ⟨Ĥ.qn⟩ in any finite number of steps.
> 

Which is a statment that means nothing, as you have implicitly agreed.

An aborted simulation does NOT indicate that the input is a non-halting 
machine.

By the exact same logic, a machine that simulated ONE step and then 
aborted can say that it was unable to reach the final state.

Remember, in all the machine you looked at, each H^ is only looked at by 
ONE decider, which simulated for some number of steps and then gave up 
and aborted.

NONE of the other machine/input pairs have ANY connection to that input, 
as they are simulations of DIFFERENT inputs.

You may be figuring out how to word your statements to get the answer 
you want, but in doing so you are removing all the meaning out of that 
statement that you need to go on.

All you have shown is that for this type of input, you simulator aborts 
its simulation before it gets to the end, and tries to find logic, which 
doesn't exist, to say it was right in aborting.

Since for every input, based on a simulator that does decide to abort 
and return 0, we can simulate that input with a UTM (or a variation of H 
that just simulates long enough) and reach the final state, your claim 
that embedded_H "can't" reach the final state is incorrect. It DOESN'T 
reach the final state, but for the input it was given, if it simulated 
longer, it would have. Note the key, the input it was given, as that is 
the problem in view, the specific input given to the specific H you want 
to present as your correct halt decider.