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

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Sujet : Re: Two dozen people were simply wrong --- Try to prove otherwise
De : mikko.levanto (at) *nospam* iki.fi (Mikko)
Groupes : comp.theory
Date : 03. Jun 2024, 11:03:56
Autres entêtes
Organisation : -
Message-ID : <v3k4ec$3ri9o$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
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

Date Sujet#  Auteur
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