Sujet : Re: Two dozen people were simply wrong --- Try to prove otherwise
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory sci.logicDate : 31. May 2024, 02:54:52
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v3bale$222n5$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
User-Agent : Mozilla Thunderbird
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.
>
The maning of x is that there is a universal
Turing machine that, when given x and y, simulates what the described
Turing machine does when given y.
>
Yes that is also correct.
>
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 then it never halts.
But it isn't unless H is also a UTM, and then H never returns.
You like to keep returning to that deception.
>
When embedded_H is a simulating halt decider then its correctly
simulated input never reaches its own simulated final state of
⟨Ĥ.qn⟩ and halts. H itself does halt and correctly rejects its
input as non-halting.
Except that isn't what the question is, the question is what the actual behavior of the machine described, or equivalently, the simulation by a REAL UTM (one that never stops till done).
When embedded_H is a real UTM then Ĥ ⟨Ĥ⟩ never stops and embedded_H is
not a decider.
When embedded_H is based on a real UTM then ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated
by embedded_H never reaches its own simulated final state of ⟨Ĥ.qn⟩ in
any finite number of steps and after these finite steps embedded_H
halts.
*I am going to stop here and not respond to anything else*
*that you say until AFTER this one point is fully resolved*
-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer