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

On 2025-03-17 13:24:24 +0000, olcott said:

On 3/17/2025 4:08 AM, Mikko wrote:

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)).

 ?- LP = not(true(LP)).
LP = not(true(LP)).

Meaning that LP = not(true(LP)) is accepted as a valid query and evalated
as true with the implication that LP is the same as not(true(LP)).

?- unify_with_occurs_check(LP, not(true(LP))).
false.

Meaning that unify_with_occurs_check(LP, not(true(LP))) is accepted as a
valid query and evaluated as false.

(SWI-Prolog (threaded, 64 bits, version 7.6.4)
 BEGIN:(Clocksin & Mellish 2003:254)
Finally, a note about how Prolog matching sometimes differs from the unification used in Resolution. Most Prolog systems will allow you to satisfy goals like:
   equal(X, X).
   ?- equal(foo(Y), Y).
 that is, they will allow you to match a term against an uninstantiated subterm of itself. In this example, foo(Y) is matched against Y, which appears within it. As a result, Y will stand for foo(Y), which is foo(foo(Y)) (because of what Y stands for), which is foo(foo(foo(Y))), and soon. So Y ends up standing for some kind of infinite structure.
END:(Clocksin & Mellish 2003:254)

The text merely says that the unification of Prolog is not always what
one might expect.
The quote does not contradict my comment that Prolog does not prove.
In particular, the ansers to your example queries were not proofs.
--
Mikko