Re: H ⟨Ĥ⟩ ⟨Ĥ⟩ is correct when reports on the actual behavior that it sees --outermost H--

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.