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

On 3/15/2025 9:18 PM, dbush wrote:

On 3/15/2025 9:47 PM, olcott wrote:

On 3/15/2025 8:27 PM, dbush wrote:

On 3/15/2025 9:03 PM, olcott wrote:

On 3/15/2025 2:25 PM, dbush wrote:

On 3/15/2025 3:16 PM, olcott wrote:

On 3/15/2025 2:05 PM, dbush wrote:

On 3/15/2025 2:55 PM, olcott wrote:

On 3/15/2025 12:39 PM, dbush wrote:

On 3/15/2025 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.
>

>
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.

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

>
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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That doesn't change the fact that

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You have just proven you are clueless about these things
by your next statement.
>

that ~True(LP) evaluates to true.
>

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When
LP  := ~True(LP)  True_Bearer(LP) == FALSE

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And by the above function, because True_Bearer(LP) == FALSE:
>

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Which means that LP cannot possibly be either TRUE or FALSE
and instead must be rejected as invalid input to a True(X)
predicate.

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False.  The True() predicate must return "true" for true statements and false for *all other statements*.
>
The fact that the True() you've defined *does* accept non-truth bearers and returns false for them shows you know this, but are being deliberately deceptive.
>

>

True(LP) == FALSE, then
~True(LP) == TRUE, so
LP == TRUE
>
Contradiction.  Therefore the assumption that a correct True() predicate exists is proven false
>

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Likewise Truth_Bearer("ksdnf34589jknsdf34r87&%78^%78") == FALSE
<sarcasm>
and on that basis we know that True(X) predicates cannot
exist because True(X) predicates must correctly determine
whether random gibberish is true or false.
</sarcasm>
>

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True(X) predicates must correctly determine
whether random gibberish is true or *not true*.  And because random gibberish is not true, True("ksdnf34589jknsdf34r87&%78^%78") must return false
>

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That is fine, and makes Tarski wrong.
>

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Nope.  Tarski uses a proof by contradiction.  You know, that type of proof you still don't understand 50 years after learning it.
>
He starts by assuming that a True() predicate exists in a system that can express the full properties of natural numbers.
>
He then shows that it's possible to create in the system that can be shown in a meta system to effectively mean:
>
LP  := ~True(LP)
>
Given that True(LP) == false, we then have ~True(LP) == true.  And since ~True(LP) is the same as LP, that gives us LP == true.
>
Contradiction.

>
True(LP) == FALSE

 And ~True(LP) == TRUE
Therefore LP == TRUE
 Contradiction.
 Therefore the assumption that a True() predicate exists is proven false

NO STUPID THAT IS NOT IT !!!
True(LP) == FALSE and True(~LP) == FALSE
thus tossing LP out on its ass for being semantically unsound.
Don't f-cking erase the context that you are responding to
you damned liar.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer