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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 12:18:54 -0500
Organization: A noiseless patient Spider
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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]

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