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On 3/20/2025 7:57 AM, Mikko wrote:Of cours. But that is irrelevant to the fact that you quoted a textOn 2025-03-17 13:18:42 +0000, olcott said:If ?- equal(foo(Y), Y)
On 3/17/2025 4:04 AM, Mikko wrote:Doesn't matter. Hopefully readers can see that you are dishonest butOn 2025-03-16 14:38:16 +0000, olcott said:*It seems to me that you are dishonest abut that*
On 3/16/2025 8:19 AM, Mikko wrote:They don't say so in the above quoted text. What they do say is essentiallyOn 2025-03-15 17:15:39 +0000, olcott said:LP := ~True(LP) expanded to infinite recursion
On 3/11/2025 5:50 AM, Mikko wrote:The above is irrelevant to the fact that you didn't say anothing aboutOn 2025-03-11 03:23:51 +0000, olcott said:I said that unify_with_occurs_check() detects
On 3/10/2025 9:49 PM, dbush wrote:That you can quote some text but don't say anything about it supports theOn 3/10/2025 10:39 PM, olcott wrote:It may seem that way if you fail to understandOn 3/10/2025 9:21 PM, Richard Damon wrote:~True(LP) resolves to trueOn 3/10/25 9:45 PM, olcott wrote:bool True(X)On 3/10/2025 5:45 PM, Richard Damon wrote:But is irrelevent to your arguement.On 3/9/25 11:39 PM, olcott wrote:The Liar Paradox PROPERLY FORMALIZED <is> Infinitely recursiveLP := ~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.
thus semantically incorrect.
"It would then be possible to reconstruct the antinomy of the liarRight, the "Liar" is in the METALANGUAGE, not the LANGUAGE where the predicate is defined.
in the metalanguage, by forming in the language itself a sentence"
You are just showing you don't understand the concept of Metalanguage.
Thus anchoring his whole proof in the Liar Paradox even ifYes, 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.
you do not understand the term "metalanguage" well enough
to know this.
{
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.
Clocksin & Mellish explanation of
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.
ON PAGE 3
https://www.researchgate.net/ publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence
hypthesis that you don't understand the text you quoted.
cycles in the directed graph of the evaluation
sequence of an expression that does explain
everything even if it seems like I said
blah, blah, blah to everyone not knowing the
meaning of these words: "cycle", directed graph"
"evaluation sequence".
the text you quoted.
~True(~True(~True(~True(~True(~True(...))))))
The same way that Clocksin and Mellish do on their example
that you dishonestly keep ignoring.
what I have said in another context but not relevant here.
that is their problem, not yours or mine.
BEGIN:(Clocksin & Mellish 2003:254)The above quote is irrelevant to the question whether ~True(LP) resolves
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)
to true.
resolves to foo(foo(foo(foo(foo(foo(...))))))
then ?- LP = not(true(LP)).
resolves to not(true(not(true(not(true(not(true(...))))))))
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