Sujet : Re: Two dozen people were simply wrong --- Try to prove otherwise
De : mikko.levanto (at) *nospam* iki.fi (Mikko)
Groupes : comp.theoryDate : 03. Jun 2024, 11:03:56
Autres entêtes
Organisation : -
Message-ID : <v3k4ec$3ri9o$1@dont-email.me>
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User-Agent : Unison/2.2
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:
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:
x <is> a finite string Turing machine description that SPECIFIES behavior. The term: "representing" is inaccurate.
No, x is a description of the Turing machine that specifies the behaviour
that H is required to report.
That is what I said.
Note, the string doesn't DIRECTLY specify behavior, but only indirectly as a description/representation of the Turing Mach
The string directly SPECIFIES behavior to a UTM or to
any TM based on a UTM.
An UTM interpretes the string as a specification of behaviour
YES, exactly !!!
and another Turing machine may interprete likewise. But in a
different context the interpretation is different.
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
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.
-- Mikko