Re: H(D,D)==0 is correct when reports on the actual behavior that it sees --outermost H--

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

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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.

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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).

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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.
>

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What does "some H(D,D)" mean? There is only one H(D,D).

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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.
>

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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.

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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