Re: Olcott is correct on this point

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Sujet : Re: Olcott is correct on this point
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theory
Date : 15. Jun 2025, 19:38:16
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <e0d869d3e181585d179192fface10cd77564f249@i2pn2.org>
References : 1 2 3
User-Agent : Mozilla Thunderbird
On 6/15/25 10:31 AM, olcott wrote:
On 6/15/2025 3:50 AM, Mikko wrote:
On 2025-06-14 15:24:58 +0000, Mr Flibble said:
>
A halting decider cannot and should not report on the behaviour of its
caller.
>
Worng.
 A partial halt decider is only allowed to report on the
behavior specified by the sequence of state transitions
of its input.
But inputs are not a "sequence of state transistions", that more describes the output of a simulator.
The input is a specification of the algorithm and data that algorithm will be applied to.

 int sum(int x, int y) { return x + y; }
sum(3,2) is not allowed to report on sum(5,7).
Right, and H(D) isn't allowed to report on the behavior of hypothetical_D that is based on Hypothetical_H.

 
The exitence of the caller and the identity if one exists are not
even mentioned in the halting problem.
 Because no one ever noticed that it is impossible
to define *AN ACTUAL INPUT* that *ACTUALLY DOES* the
opposite of whatever its value its corresponding
partial halt decider returns.
Sure it is.

 int main()
{
   DDD(); // calls HHH(DDD) that does not report on
}        // the behavior of its caller.
And that just makes it wrong, since DDD is the specification of the caller to HHH, so that is what HHH needs to report on.

 When Ĥ is applied to ⟨Ĥ⟩     // Peter Linz Proof.
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
 embedded_H does not report on the behavior of the
computation that its actual self is contained within.
But it must, since (H^) (H^) represent the computation it is contained in.
I guess you are just admitting that you whole proof is just a lie.

 ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H cannot possibly
reach its own simulated final halt state of ⟨Ĥ.qn⟩
But that presuposses the LIE that embedded_H does correctly simulate its input.
If the definition of H is such that this is a true statement, then (as you have shown) H (and embedded_H) can't ever return an answer, and thus H isn't a decider.

 Only because I have spent 22 years on this have I
noticed details that no one else has ever noticed before.
No, you have spent 22 years lying to yourself and believing those lies.
The problem is you chose to remain ignorant of the basics of the field, so you could keep trying to believe your own lies, but all that has done is turn your into a pathological liar that is just too stupid to understand his error.

 
THerefore they don't affect what
a halting decider or a partial halting report is required to report.
>
There are partial halting deciders that can correctly report on the
behaviours of some of their callers.
>
 

Date Sujet#  Auteur
15 Jun 25 * Re: Olcott is correct on this point4Mikko
15 Jun 25 `* Re: Olcott is correct on this point3olcott
15 Jun 25  +- Re: Olcott is correct on this point1Richard Damon
16 Jun 25  `- Re: Olcott is correct on this point1Mikko

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