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

On 2025-03-15 17:08:33 +0000, olcott said:

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.

Prolog does not prove anything other than what you ask. I don't think
you can ask Prolog whether there is a cycle in LP after LP = not(true(LP)).
--
Mikko