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On 7/19/2025 12:23 PM, Richard Damon wrote:That your are quoting from never-learned ignorance doesn't releave you of the sin of lying, since it has been explained to you what the true definition is,On 7/19/25 10:23 AM, olcott wrote:>>If you only learn by rote memorization it may seem that way.When you change my words and use those words as the basis>
of your rebuttal you know that you cheat.
>
The infinite simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H
cannot possibly reach its own simulated final
halt state of ⟨Ĥ.qn⟩ you fucking moron.
Which is just a strawman, as the definition of non-halting behavior of the input, is that when the machine the input describes,
When you have a depth of understanding then you see that
what you said is merely an incorrect paraphrase of this:
Turing machine (at least partial) halt deciders only computeAnd that behavior is DEFINED by the behavior of the direct exectution of the program it represents (and it must represent a program, including ALL the code it uses).
the mapping from their finite string inputs to the actual
behavior that this input finite string actually specifies.
Conventional notation of a Turing Machine: ĤWRONG, READ THE SPECS AGAIN.
Conventional notation of a TM description: ⟨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.
*Is corrected to this*
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H reaches
its simulated final halt state of ⟨Ĥ.qn⟩, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H cannot possibly
reach its simulated final halt state of ⟨Ĥ.qn⟩.
*Original proof*Just rote repeating your lies just proves you are a pathological liar.
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
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