Re: How do we know that ChatGPT 4.0 correctly evaluated my ideas?

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Sujet : Re: How do we know that ChatGPT 4.0 correctly evaluated my ideas?
De : news (at) *nospam* immibis.com (immibis)
Groupes : comp.theory
Date : 07. Mar 2024, 17:50:35
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <uscnob$14dmi$2@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 25 26 27 28 29
User-Agent : Mozilla Thunderbird
On 7/03/24 16:38, olcott wrote:
On 3/7/2024 5:44 AM, Mikko wrote:
On 2024-03-06 17:08:25 +0000, olcott said:
>
On 3/6/2024 3:06 AM, Mikko wrote:
On 2024-03-06 07:11:34 +0000, olcott said:
>
Chat GPT CAN'T understand the words, it has no programming about MEANING.
>
You cant find any mistakes in any of its reasoning.
>
*This paragraph precisely follows from its preceding dialogue*
>
When an input, such as the halting problem's pathological input D, is
designed to contradict every value that the halting decider H returns,
it creates a self-referential paradox that prevents H from providing a
consistent and correct response. In this context, D can be seen as
posing an incorrect question to H, as its contradictory nature
undermines the possibility of a meaningful and accurate answer.
>
That is essentially an agreement with Linz proof.
*It is not an agreement with the conclusion of this proof*
>
Not explicitly but comes close enough that the final step is
trivial.
>
It is an agreement with why Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer.
>
That, too.
>
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
 The Linz proof correctly proves that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
can't possibly get the right answer and falsely
concludes that this means that H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot
get the correct answer.
 *My H(D,D) and H1(D,D) prove otherwise*
 
An embedded copy of a machine is stipulated to always get the same result as the original machine. If this is not true, then you did not make an embedded copy. Making an embedded copy of a Turing machine is straightforward. Making an embedded copy of an Olcott machine is a bit more tricky - you have to make sure that whenever the original accesses "my own description", the embedded copy includes a copy of whatever that description would be for the original machine, and accesses that instead.

Date Sujet#  Auteur
21 Sep 24 o 

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