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From: olcott <polcott2@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
 sees --outermost H--
Date: Fri, 15 Mar 2024 12:57:18 -0500
Organization: A noiseless patient Spider
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On 3/15/2024 12:39 PM, immibis wrote:
> On 15/03/24 18:18, olcott wrote:
>> On 3/15/2024 12:15 PM, immibis wrote:
>>> On 15/03/24 18:11, olcott wrote:
>>>> On 3/15/2024 12:06 PM, immibis wrote:
>>>>> On 15/03/24 15:17, olcott wrote:
>>>>>> On 3/15/2024 4:36 AM, Fred. Zwarts wrote:
>>>>>>> Op 15.mrt.2024 om 03:40 schreef olcott:
>>>>>>>> On 3/14/2024 9:34 PM, immibis wrote:
>>>>>>>>> On 15/03/24 03:29, olcott wrote:
>>>>>>>>>>
>>>>>>>>>> *Actually it is the fact that the top H ⟨Ĥ⟩ ⟨Ĥ⟩ (not a copy) 
>>>>>>>>>> does*
>>>>>>>>>> *get this correctly that proves that H ⟨Ĥ⟩ ⟨Ĥ⟩ does not meet the*
>>>>>>>>>> *original criteria because it does meet the above criteria*
>>>>>>>>>>
>>>>>>>>>> Execution trace of H applied to ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>>>>>>> (1) H applied ⟨Ĥ⟩ ⟨Ĥ⟩ simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
>>>>>>>>>> (2) which begins at simulated ⟨Ĥ.q0⟩
>>>>>>>>>> (a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
>>>>>>>>>> (b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied 
>>>>>>>>>> to ⟨Ĥ⟩
>>>>>>>>>> (c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the 
>>>>>>>>>> process
>>>>>>>>>>
>>>>>>>>>> The earliest point when Turing machine H can detect the repeating
>>>>>>>>>
>>>>>>>>> Whensoever H detects the repeating state and aborts it is 
>>>>>>>>> incorrect because the state is not repeating. The state is 
>>>>>>>>> repeating if H does not detect the repeating state.
>>>>>>>>
>>>>>>>> You keep saying that H(D,D) never really needs to abort the
>>>>>>>> simulation of its input because after H(D,D) has aborted the
>>>>>>>> simulation of this input it no longer needs to be aborted.
>>>>>>>>
>>>>>>>
>>>>>>> Do you finally understand it? Hah(Dah,Dah) does not need to 
>>>>>>> abort, because Dah halts. Hah should look at its input Dah (which 
>>>>>>> aborts), not at its non-input Dss (which does not abort).
>>>>>>
>>>>>> Unless some H(D,D) aborts the simulation of its input D(D) never 
>>>>>> stops
>>>>>> running. The outermost H(D,D) sees this abort criteria first. If the
>>>>>> outermost H(D,D) does not abort its simulation then none of them do.
>>>>>> therefore the outermost H(D,D) is correct to abort its simulation.
>>>>>>
>>>>>
>>>>> What does "some H(D,D)" mean? There is only one H(D,D).
>>>>
>>>> D(D) specifies an infinite chain of H(D,D) unless D(D) is aborted
>>>> at some point. The outermost H(D,D) always has seen a longer execution
>>>> trace than any of the inner ones.
>>>>
>>>
>>> D(D) only specifies one call to H(D,D). It is H's fault if H is 
>>> unable to return a value without infinite recursion.
>>
>> This conversation has been moved to here:
>> [Proof that H(D,D) meets its abort criteria]
>>
> 
> Strawman deflection ignored. D(D) only specifies one call to H(D,D). It 
> is H's fault if H is unable to return a value without infinite recursion.

This conversation has been moved to here:
[Proof that H(D,D) meets its abort criteria]

After we have 100% complete closure on that point
then we can change back to H ⟨Ĥ⟩ ⟨Ĥ⟩.

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer