Re: My reviewers think that halt deciders must report on the behavior of their caller

Op 11.jul.2025 om 23:05 schreef olcott:

On 7/11/2025 3:52 AM, Fred. Zwarts wrote:

Op 10.jul.2025 om 16:35 schreef olcott:

On 7/10/2025 5:54 AM, Fred. Zwarts wrote:

Op 09.jul.2025 om 15:02 schreef olcott:>

All Turing machine deciders only compute the mapping
from their actual inputs. This entails that they never
compute any mapping from non-inputs.

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At least one thing you understand.
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*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
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*The Linz proof does not understand this*
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When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
   *if Ĥ applied to ⟨Ĥ⟩ halts, and*
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
   *if Ĥ applied to ⟨Ĥ⟩ does not halt*
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The evidence is that the input includes the code to abort and halt,

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abort and stop running
*IS NOT THE SAME THING AS*
abort and halt

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Another claim without evidence.
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*It is common knowledge in the theory of computation*
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Another claim without evidence.
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 *Your lack of knowledge of computer science is not a rebuttal*
 Look at the definition of a Turing Machine (e.g., the one here). The machine has states. Each state can be final or non-final. If the machine's state is non-final, in the next step the machine "does" something, namely, it can write something on the tape, move its head, and/or change its state to a different state. This is how the machine makes a progress.

So, aborting the simulation when the machine has not yet reached its final state, is a violation of the Turing Machine.

 However, if the state is final, the machine just stops and continues no more. This will be its final setting, the head will not move anymore and the tape remains the same forever. This event is non-reversible and is what we call halting.

Exactly. And halting is what DDD does, when not disturbed by a premature abort that violates the semantics of the x86 language.