Re: Turing Machine computable functions apply finite string transformations to inputs VERIFIED FACT

On 4/30/25 10:58 AM, olcott wrote:

On 4/30/2025 1:05 AM, Richard Heathfield wrote:

On 30/04/2025 03:45, Richard Damon wrote:

On 4/29/25 3:57 PM, olcott wrote:

On 4/29/2025 10:33 AM, Fred. Zwarts wrote:

Op 29.apr.2025 om 15:11 schreef olcott:

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No H can possibly see the behavior of P(D)
when-so-ever D has defined a pathological
relationship with H this

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makes it impossible for H to see the behaviour of P(D).
The behaviour of P(D) does not change, but H does not see it.

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H MUST REPORT ON THE BEHAVIOR THAT IT DOES SEE
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No, it must report on the behavior that exists.
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It is only ABLE to correctly report on behavior it can "see", but there is no structural restriction that says we can't ask it about something that it can't see.

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Nor is there any restriction that says it can't deduce behaviour it can't see, simply by reading the tapes.
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 It is ONLY allowed to derive its output by
applying finite string transformations to its input.
For the HHH/DD pair these must be the finite string
transformations specified by the x86 language.]

It is only ABLE to.
The correct answer comes from the Halting Function mapping which has no such restriction.

 For the embedded_H / ⟨Ĥ⟩ ⟨Ĥ⟩ pair these
transformations are specified by the Linz template.
 When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
 Above adapted from bottom of page 2
https://www.liarparadox.org/Linz_Proof.pdf
 (a) Ĥ copies its input ⟨Ĥ⟩
(b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
(e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(g) goto (d) with one more level of simulation

And since H (Ĥ) (Ĥ) aborts its emulation and return to qn, then so must the embedded_H started in (c), and when that happens we end up unwinding   and then going do (d1) where embedded_H goes to Ĥ.qn and Ĥ halts.

 

Mr Olcott seems unable to recognise this possibility. Having built his hammer, he is determined to see the Halting Problem as a nail that cannot withstand being pounded hard enough.
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Unfortunately for him, the problem is more like a 16 puzzle with two tiles swapped. If he plays by the rules there is no solution, no matter how hard he hits it.
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This is why he keeps trying to change the rules, with word salad like 'pathological self-reference', when self-reference is the whole reason the proof works.
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