Sujet : Re: The key undecidable instance that I know about --- Truth-bearers ONLY
De : dbush.mobile (at) *nospam* gmail.com (dbush)
Groupes : sci.logicDate : 16. Mar 2025, 03:41:21
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vr5dog$pp4d$5@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
User-Agent : Mozilla Thunderbird
On 3/15/2025 10:27 PM, olcott wrote:
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.
>
>
The Liar Paradox PROPERLY FORMALIZED <is> Infinitely recursive
thus semantically incorrect.
>
But is irrelevent to your arguement.
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>
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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.
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that ~True(LP) evaluates to true.
>
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When
LP := ~True(LP) True_Bearer(LP) == FALSE
>
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.
>
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:
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LP := ~True(LP)
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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.
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Contradiction.
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True(LP) == FALSE
>
And ~True(LP) == TRUE
Therefore LP == TRUE
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Contradiction.
>
Therefore the assumption that a True() predicate exists is proven false
NO STUPID THAT IS NOT IT !!!
In other words, you're once again demonstrating that you don't understand proof by contradiction, a concept taught to and understood by high school students more than 50 years your junior.
The key undecidable instance that I know about
Re: The key undecidable instance that I know about