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From: immibis <news@immibis.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 18:39:07 +0100
Organization: A noiseless patient Spider
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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.