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From: immibis <news@immibis.com>
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Date: Fri, 15 Mar 2024 17:56:08 +0100
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On 15/03/24 03:40, olcott wrote:
> 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.
> 
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.