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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: My reviewers think that halt deciders must report on the behavior
 of their caller
Date: Mon, 7 Jul 2025 18:45:52 -0400
Organization: i2pn2 (i2pn.org)
Message-ID: <2e55c544e1eafae549ef3181a20784c6a248a661@i2pn2.org>
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On 7/7/25 10:15 AM, olcott wrote:
> On 7/7/2025 3:37 AM, Mikko wrote:
>> On 2025-07-07 03:12:30 +0000, olcott said:
>>
>>> On 7/6/2025 9:09 PM, Richard Damon wrote:
>>>> On 7/6/25 4:06 PM, olcott wrote:
>>>>> On 7/6/2025 12:00 PM, Richard Damon wrote:
>>>>>> On 7/6/25 11:19 AM, olcott wrote:
>>>>>>>
>>>>>>> void DDD()
>>>>>>> {
>>>>>>>    HHH(DDD);
>>>>>>>    return;
>>>>>>> }
>>>>>>>
>>>>>>> *EVERY BOT FIGURES THIS OUT ON ITS OWN*
>>>>>>
>>>>>> No, it just isn't smart enough to detect that you lied in your 
>>>>>> premise.
>>>>>>
>>>>>>> There is no way that DDD simulated by HHH (according
>>>>>>> to the semantics of the C programming language)
>>>>>>> can possibly reach its own "return" statement final
>>>>>>> halt state.
>>>>>>
>>>>>> And there is no way for HHH to correctly simulate its input and 
>>>>>> return an answer
>>>>>>
>>>>>
>>>>> You insistence that a non-terminating input be simulated
>>>>> until non-existent completion is especially nuts because
>>>>> you have been told about this dozens of times.
>>>>>
>>>>> What the F is wrong with you?
>>>>>
>>>>
>>>> It seems you don't understand those words.
>>>>
>>>> I don't say that the decider needs to simulate the input to 
>>>> completion, but that it needs to be able to actually PROVE that if 
>>>> this exact input WAS given to a correct simultor (which won't be 
>>>> itself, since it isn't doing the complete simulation) will run for 
>>>> an unbounded number of steps.
>>>
>>> No decider is ever allowed to report on anything
>>> besides the actual behavior that its input actually
>>> specifies.
>>
>> Unless you can quote some respectable author your prohibitions are
>> meaningless.
>>
> 
> To people that never had any actual understanding and
> can only parrot textbooks. They need to see this things
> in other textbooks.

Better than someone who is too stupid to understand the textbooks that 
DEFINE the system, ahd thus you just prove your ignorance.

> 
> It is common knowledge that Turing Machine Halt Deciders
> can only take finite string encodings of Turing Machines
> as inputs. Thus anything that it not a finite string is
> outside of the domain of Turing Machine Halt Deciders.

Right, but since it is an "Encoding of a Turing Machine", the behavor of 
that machine is fair game to ask about.

> 
> People have always know the first sentence of that and
> never bothered to derive the second sentence from the first.

It seems you don't understand what you can do with encodings.

You don't understand that EVERYTHING you do is with an "encoding" but 
then, I don't think you understand how anything works.

> 
>>> Most people here don't get that because they have no
>>> actual depth of understanding. They can only parrot
>>> the words of textbooks.
>>
>> Do you even understand what the word "allowed" means?
>>
> Outside of the domain is a more precise way of saying it.
> 
> *From the bottom of page 319 has been adapted to this*
> https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
> 
> When Ĥ is applied to ⟨Ĥ⟩
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
>     ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H reaches
>     its simulated final halt state of ⟨Ĥ.qn⟩
> 
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>     ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H cannot possibly
>     reach its simulated final halt state of ⟨Ĥ.qn⟩
> 

But why did you need to channge H into embedded_H?

And what do H / embedded_H transition to (and they will do the same) to 
be correct.

If they go to H,qy then H^ will not halt, but qy is supposed to only be 
gone to if the input halts.

If they go to H.qn then H^ will halt, but qn is supposed to only be gone 
to if the input never halts.

Thus, you are just proving that you are a liar.