Re: Why does Olcott care about simulation, anyway?

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Sujet : Re: Why does Olcott care about simulation, anyway?
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theory sci.logic
Date : 05. Jun 2024, 03:47:59
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <v3og4f$328ec$3@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
User-Agent : Mozilla Thunderbird
On 6/4/24 1:06 PM, olcott wrote:
On 6/3/2024 9:53 PM, Richard Damon wrote:
On 6/3/24 10:47 PM, olcott wrote:
On 6/3/2024 9:26 PM, Richard Damon wrote:
On 6/3/24 10:09 PM, olcott wrote:
On 6/3/2024 8:58 PM, Richard Damon wrote:
On 6/3/24 9:54 PM, olcott wrote:
On 6/3/2024 8:44 PM, Richard Damon wrote:
On 6/3/24 9:05 PM, olcott wrote:
>
*I say that you know you are a liar until after you show the steps*
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DD will halt (Remember, I am not saying the somulaiton by HH, but that DD itself will halt).
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That IS the strawman deception that might possibly (I hope not)
get you condemned to Hell.
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What is strawman about it?
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I am just using the actual definitions that YOU like to ignore and make lies about.
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You may condemn yourself to Hell by even asking that question.
I hope not. I myself wouldn't risk it.
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I guess you think God hates people who bring out the Truth,
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Yet because you know that you keep changing the subject from DD correctly simulated by HH to the directly executed DD(DD)
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Because you keep on mentioning about DD Halting, which IS about the direct execution of DD
 Only when one contradicts the definition of a decider that must
compute the mapping FROM ITS INPUTS BASED ON THE ACTUAL BEHAVIOR
OF THESE INPUTS (as measured by DD correctly simulated by HH).
Please show a source for your LIE that the definition is based on the "correct simulation by HH"
The ACTUAL definition is as defined by the mathematical function it is trying to compute, which for a halt decider is "Halts" which returns true if the machine described by the input will halt in a finite number of steps with the provided input, and false if it will never halt even after an unbounded number of steps.

 When we go ahead and contradict this definition then the
*HALTING PROBLEM IS STILL WRONG IN A DIFFERENT WAY*
Nope. YOU are wrong.

 When D is defined to do the opposite of whatever yes/no
an answer that H provides then the counter-example input
is precisely isomorphic to the question:
Is this sentence: "This sentence is not true." true or false?
Nope, because the answer that H doesn't give is right.
There IS a correct answer for the input, it just is fact that H doesn't give it.
Somehow you think that deterministic algorithms have something like free will and can give different answer as needed, they can't. And you idea that they can says you don't understand the fundamental properties of programs.
I wonder if the issue is that you lost your free will in a deal with the devil, and you are trying to prove that you can get it back.

 The theory of computation may be ignorant of the details of
how the context of who is asked a question changes the meaning
of this question, none-the-less this cannot be ignored.
It is and remains incorrect for the theory of computation
to ignore this.
 
Contex CAN change the meaning, if the question is subjective, but deciders can only be asked OBJECTIVE questions, as that is all the mathematics allows to be expressed.
Of course, since your logic system appears not to be able to handle actual mathematics as it is limited to what Prolog can handle, so that is beyond your abilities.

Date Sujet#  Auteur
10 Nov 24 o 

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