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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory,sci.logic
Subject: Re: My reviewers think that halt deciders must report on the behavior
 of their caller
Date: Thu, 10 Jul 2025 07:36:50 -0400
Organization: i2pn2 (i2pn.org)
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On 7/9/25 9:16 AM, olcott wrote:
> On 7/9/2025 4:04 AM, Fred. Zwarts wrote:
>> Op 08.jul.2025 om 16:31 schreef olcott:
>>> On 7/8/2025 2:55 AM, Fred. Zwarts wrote:
>>>> Op 08.jul.2025 om 04:52 schreef olcott:
>>>>> On 7/7/2025 9:24 PM, Richard Damon wrote:
>>>>>> On 7/7/25 7:47 PM, olcott wrote:>>
>>>>>>> That Turing machines cannot take directly executing Turing
>>>>>>> Machines as inputs entails that these directly executed
>>>>>>> machines are outside of the domain of every Turing machine
>>>>>>> based halt decider.
>>>>>>
>>>>>> But they can take the finite-stringt encoding of those machines.
>>>>>>
>>>>>
>>>>> Yes.
>>>>>
>>>>>> I guess you idea of Turing Machine is so limited that you think 
>>>>>> they can't do arithmatic, as you can't actually put a "Number" as 
>>>>>> the input, only the finite-string encoding of a number, which puts 
>>>>>> it outside the domain of them.
>>>>>>
>>>>>
>>>>> No one here has any understanding of the philosophy of
>>>>> computation. They can only memorize the rules and have
>>>>> no idea about the reasoning behind these rules.
>>>>>
>>>>>>>
>>>>>>> That you cannot understand that is a truism is only your
>>>>>>> own lack of understanding.
>>>>>>
>>>>>> But it isn't a truism, it is just a stupid lie that ignores that 
>>>>>> almost everything done with programs is via an "encoding" for the 
>>>>>> input.
>>>>>>
>>>>>
>>>>> Gross ignorance about the reasoning behind the rules
>>>>> of computation would tell you that.
>>>>>
>>>>>>>
>>>>>>> https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
>>>>>>> *Here is the Linz proof corrected to account for that*
>>>>>>>
>>>>>>> *adapted from bottom of page 319*
>>>>>>> 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⟩
>>>>>>>
>>>>>>>
>>>>>>
>>>>>> Which is just an admission of your lying strawman, as the question 
>>>>>> is NOT about the (partial) simulation done by your H / embedded_H, 
>>>>>> but about the direct execution of the input H^ (H^) as that is 
>>>>>> what the input to H is encoding.
>>>>>>
>>>>>
>>>>> Because no Turing machine can take a directly executed
>>>>> Turing machine as an input, directly executed Turing
>>>>> machines have always been outside of the domain of every
>>>>> Turing machine based decider.
>>>>>
>>>>> "the direct execution of the input H^ (H^)" has always been
>>>>> out-of-scope for every Turing machine based halt decider.
>>>>> That no one bothered to notice this ever before
>>>>> *DOES NOT MAKE ME WRONG*
>>>> That is your misconception. No one ever asked to take the direct 
>>>> execution as input. 
>>>
>>> *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.∞
>>>    if Ĥ applied to ⟨Ĥ⟩ halts
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>    if Ĥ applied to ⟨Ĥ⟩ does not halt
>>>
>>> *The above original form of the proof*
>>> does requires Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ to report on
>>> the direct execution of Ĥ applied to ⟨Ĥ⟩ and thus
>>> not ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by Ĥ.embedded_H.
>>>
>> Your are fighting windmills. Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ is required to 
>> report on the specification of its input Ĥ. 
> 
> Very close.
> Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ is required to report
> on the behavior that its input specifies.
> 
>> This Ĥ includes the code that make it halt, 
> 
> More precisely This Ĥ includes one final halt state: Ĥ.qn
> That is unreachable when ⟨Ĥ⟩ ⟨Ĥ⟩ is correctly simulated by
> Ĥ.embedded_H.

Which is the behavior of H^ (H^).

Since your H doesn't correctly simulate the input (or H fails to answer) 
your criteria is just a self-contradictory lie.

> 
>> so the specification is a halting program.
> 
> False conclusion from false premise.

Nope, DEFINITIONS.

> 
>>  Nowhere there is the requirement that it must report on the direct 
>> execution.
> 
> *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.∞
>    if Ĥ applied to ⟨Ĥ⟩ halts, and
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>    if Ĥ applied to ⟨Ĥ⟩ does not halt.
> 
> if Ĥ applied to ⟨Ĥ⟩ halts // is the requirement
> // that it report in the direct execution.
> 
> https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
> When M is applied to WM
> q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ∞,
>     if M applied to WM halts, and
> q0 WM ⊢* Ĥq0 Wm WM ⊢* Ĥ y1 qn y2,
>     if M applied to WM does not halt.
> The above is the original textbook quote.
> 
>> Direct execution is only one way to prove what is specified in the 
>> input, but often there are other methods to prove it. Ĥ does not need 
>> to know it, it show only report what is specified. 
> 
> I agree with you and everyone else seems to disagree with you.
> 
>> If it fails to report that a halting program halts, it is just wrong.
> 
> The input specifies recursive simulation that
> never halts even though the program halts.
> 
> This can best be seen at the x86 machine code level.
> If you have no understanding of the x86 language you
> won't be able to see this.
> 
>