Re: The key undecidable instance that I know about --- Truth-bearers ONLY

On 3/15/25 10:39 PM, olcott wrote:

On 3/15/2025 9:13 PM, Richard Damon wrote:

On 3/15/25 9:22 PM, olcott wrote:

On 3/15/2025 3:44 PM, Richard Damon wrote:

On 3/15/25 1:08 PM, olcott wrote:

On 3/10/2025 9:49 PM, dbush wrote:

On 3/10/2025 10:39 PM, olcott wrote:

On 3/10/2025 9:21 PM, Richard Damon wrote:

On 3/10/25 9:45 PM, olcott wrote:

On 3/10/2025 5:45 PM, Richard Damon wrote:

On 3/9/25 11:39 PM, olcott wrote:

>
LP := ~True(LP)  DOES SPECIFY INFINITE RECURSION.

>
WHich is irrelevent, as that isn't the statement in view, only what could be shown to be a meaning of the actual statement.
>

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The Liar Paradox PROPERLY FORMALIZED <is> Infinitely recursive
thus semantically incorrect.

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But is irrelevent to your arguement.
>
>

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"It would then be possible to reconstruct the antinomy of the liar
  in the metalanguage, by forming in the language itself a sentence"

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Right, the "Liar" is in the METALANGUAGE, not the LANGUAGE where the predicate is defined.
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You are just showing you don't understand the concept of Metalanguage.
>

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Thus anchoring his whole proof in the Liar Paradox even if
you do not understand the term "metalanguage" well enough
to know this.

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Yes, there is a connection to the liar's paradox, and that is that he shows that the presumed existance of a Truth Predicate forces the logic system to have to resolve the liar's paradox.
>

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bool True(X)
{
   if (~unify_with_occurs_check(X))
     return false;
   else if (~Truth_Bearer(X))
    return false;
   else
    return IsTrue(X);
}
>
LP := ~True(LP)
True(LP) resolves to false.

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~True(LP) resolves to true
LP := ~True(LP) resolves to true
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Therefore the assumption that a correct True() predicate exists is proven false.

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When you stupidly ignore Prolog and MTT that
both prove there is a cycle in the directed graph
of their evaluation sequence. If you have no idea
what "cycle", "directed graph" and "evaluation sequence"
means then this mistake is easy to make.
>

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WHen you claim that Prolog gives answers for logic system more advanced then it, or make unsupported claims about your FRAUD of MTT, you are just showing your stupidity.
>
Part of your problem, it seems, is that you don't understand the limitations of Prolog, because you can't understand the logic that Prolog can't handle, because you are just too stupid.

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LP := ~True(LP
Try to explain in your own words what this means:
LP specifies a cycle in the directed graph of its evaluation sequence.
>

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The problem is that LP := ~True(LP) is just an approximation in representation of the actual statement, since you can understand what x is in the language.
>

 "This sentence is not True" <is> LP := ~True(LP)
Tarski got this WRONG
 

Why do you say that, where did he say he was trying to do what you claim he was saying?
He said he go the antinomy of the liar, that is, he got something with the same form of contradiction as the liar paradox, not that he got exactly the liar paradox.
Your problem is you are too stupid to understand what is being talked about, and just assume that they must be just as stupid as you and not expressing themselves well.
In reality, all you are doing is showing your own stupidity.
Sorry, but that is the FACTS, a concept that seems foreign to you.