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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory,sci.logic
Subject: Re: Two dozen people were simply wrong --- Try to prove otherwise
Date: Thu, 30 May 2024 22:15:30 -0400
Organization: i2pn2 (i2pn.org)
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On 5/30/24 9:54 PM, olcott wrote:
> On 5/30/2024 8:37 PM, Richard Damon wrote:
>> On 5/30/24 9:31 AM, olcott wrote:
>>> On 5/30/2024 2:40 AM, Mikko wrote:
>>>> On 2024-05-30 01:15:21 +0000, olcott said:
> 
>>>>> x <is> a finite string Turing machine description that SPECIFIES 
>>>>> behavior. The term: "representing" is inaccurate.
>>>>
>>>> No, x is a description of the Turing machine that specifies the 
>>>> behaviour
>>>> that H is required to report. 
>>>
>>> That is what I said.
>>
>> Note, the string doesn't DIRECTLY specify behavior, but only 
>> indirectly as a description/representation of the Turing Mach
>>
> 
> The string directly SPECIFIES behavior to a UTM or to
> any TM based on a UTM.

By telling that UTM information about the state-transition table of the 
machine.

Note, the description of the machine doesn't depend on the input given 
to it, so it needs to fully specify how to recreate the behavior of the 
machine for ALL inputs (an infinite number of them) in a finite string.

> 
>>>
>>>> The maning of x is that there is a universal
>>>> Turing machine that, when given x and y, simulates what the described
>>>> Turing machine does when given y. 
>>>
>>> Yes that is also correct.
>>
>>
>>
>>>
>>> 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 embedded_H is a UTM then it never halts.
>>
>> But it isn't unless H is also a UTM, and then H never returns.
>>
>> You like to keep returning to that deception.
>>
>>>
>>> When embedded_H is a simulating halt decider then its correctly
>>> simulated input never reaches its own simulated final state of
>>> ⟨Ĥ.qn⟩ and halts. H itself does halt and correctly rejects its
>>> input as non-halting.
>>
>> Except that isn't what the question is, the question is what the 
>> actual behavior of the machine described, or equivalently, the 
>> simulation by a REAL UTM (one that never stops till done).
> 
> When embedded_H is a real UTM then Ĥ ⟨Ĥ⟩ never stops and embedded_H is
> not a decider.

Right, that is YOUR delema. You can't make H / embedded_H a UTM without 
making it not a decider, thus "Correct Simulation by H" can't be the 
answer, since H can't do both.

> 
> When embedded_H is based on a real UTM then ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated
> by embedded_H never reaches its own simulated final state of ⟨Ĥ.qn⟩ in
> any finite number of steps and after these finite steps embedded_H
> halts.

Then its simulation isn't "correct" per the definitions that relate 
simulation to behavior.

> 
> *I am going to stop here and not respond to anything else*
> *that you say until AFTER this one point is fully resolved*
> 

And you need to understand that if you change the UTM to abort its 
simulation, it is no longer a UTM and has lost that property that its 
simulation reveals the behavior of the input.