Re: Two dozen people were simply wrong --- Try to prove otherwise

On 6/3/2024 5:03 AM, Mikko wrote:

On 2024-05-31 15:13:02 +0000, olcott said:
 

On 5/31/2024 3:30 AM, Mikko wrote:

On 2024-05-31 01:54:52 +0000, olcott said:
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On 5/30/2024 8:37 PM, Richard Damon wrote:

On 5/30/24 9:31 AM, olcott wrote:

On 5/30/2024 2:40 AM, Mikko wrote:

On 2024-05-30 01:15:21 +0000, olcott said:

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x <is> a finite string Turing machine description that SPECIFIES behavior. The term: "representing" is inaccurate.

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No, x is a description of the Turing machine that specifies the behaviour
that H is required to report.

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That is what I said.

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Note, the string doesn't DIRECTLY specify behavior, but only indirectly as a description/representation of the Turing Mach
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The string directly SPECIFIES behavior to a UTM or to
any TM based on a UTM.

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An UTM interpretes the string as a specification of behaviour

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YES, exactly !!!
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and another Turing machine may interprete likewise. But in a
different context the interpretation is different.
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When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
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When embedded_H <is> a UTM or <is> a halting computation based on a
UTM then the ⟨Ĥ⟩ ⟨Ĥ⟩ input to embedded_H SPECIFIES that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly
simulated by embedded_H cannot possibly reach its own simulated final
state at ⟨Ĥ.qn⟩.

 There is no requirement to follow the pecifications by embedded_H.

In other words embedded_H can simply play bingo and never halt and still correctly decide the halt status of its input?
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer