Liste des Groupes | Revenir à s logic |
On 3/15/2025 2:55 PM, olcott wrote:Which means that LP cannot possibly be either TRUE or FALSEOn 3/15/2025 12:39 PM, dbush wrote:And by the above function, because True_Bearer(LP) == FALSE: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.
>
>>>
"It would then be possible to reconstruct the antinomy of the liar
in the metalanguage, by forming in the language itself a sentence"
Right, the "Liar" is in the METALANGUAGE, not the LANGUAGE where the predicate is defined.
>
You are just showing you don't understand the concept of Metalanguage.
>>>
Thus anchoring his whole proof in the Liar Paradox even if
you do not understand the term "metalanguage" well enough
to know this.
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.
>
That doesn't change the fact that
You have just proven you are clueless about these things
by your next statement.
>that ~True(LP) evaluates to true.>
>
When
LP := ~True(LP) True_Bearer(LP) == FALSE
True(LP) == FALSE, thenLikewise Truth_Bearer("ksdnf34589jknsdf34r87&%78^%78") == FALSE
~True(LP) == TRUE, so
LP == TRUE
Contradiction. Therefore the assumption that a correct True() predicate exists is proven false
Les messages affichés proviennent d'usenet.