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

On 3/15/24 10:11 AM, 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:

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*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.
>

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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).

 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.
 

Right, *AT SOME POINT* but not nessesarily HERE.
No, the outermose has seen more execution trace then the innerones have at the point that H aborts their simulation.
The actual processing of those computations continue on until then make the exact same decision (since they have the exact same algorithm and input data).
You don't seem to understand that Turing Machines are not "Programs" that run on some other piece of hardware or only exist in their simulation, but are DEFINED to be the FULL COMPUTATIONAL ENGINE, that we might observe through the simulation, but actually always run to completion, since Turing Machines have no "off" or "break" switch to stop them, nor do they need "External Power' to run.
Its sort of like you can't make some Natural Numbers "go away" by trying to abort the definition that every Natural Number has a successor.
You can't "stop" logic, it just happens. And Turing Machines are a form of LOGIC, defining their RULES that they WILL follow.