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 false.
X = "A box of rocks"
True(X) resolves to false.
True(~X) resolves to false.
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.
Right, the "Liar" is in the METALANGUAGE, not the LANGUAGE where the
"It would then be possible to reconstruct the antinomy of the liar
in the metalanguage, by forming in the language itself a sentence" >>>>
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
It may seem that way if you fail to understand
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
On 3/11/2025 5:50 AM, Mikko wrote:
On 2025-03-11 03:23:51 +0000, olcott said:
It may seem that way if you fail to understand
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
That you can quote some text but don't say anything about it supports the
hypthesis that you don't understand the text you quoted.
*That you keep ignoring it is either stupid or dishonest*
As a result, Y will stand forfoo(Y), which is
foo(foo(Y)) (because of what Y stands for),
which is foo(foo(foo(Y))), and so on. So Y ends
up standing for some kind of infinite structure.
On 3/11/2025 6:12 AM, Richard Damon wrote:
On 3/10/25 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.
Right, the "Liar" is in the METALANGUAGE, not the LANGUAGE where the
"It would then be possible to reconstruct the antinomy of the liar
in the metalanguage, by forming in the language itself a sentence" >>>>
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 false.
X = "A box of rocks"
True(X) resolves to false.
True(~X) resolves to false.
Not the system that Tarski is using.
*That is 100% of his complete error*
On 3/11/2025 5:50 AM, Mikko wrote:
On 2025-03-11 03:23:51 +0000, olcott said:
It may seem that way if you fail to understand
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
That you can quote some text but don't say anything about it supports the
hypthesis that you don't understand the text you quoted.
*That you keep ignoring it is either stupid or dishonest*
As a result, Y will stand forfoo(Y), which is
foo(foo(Y)) (because of what Y stands for),
which is foo(foo(foo(Y))), and so on. So Y ends
up standing for some kind of infinite structure.
On 3/12/2025 5:39 AM, Mikko wrote:
On 2025-03-11 23:10:14 +0000, olcott said:
On 3/11/2025 5:50 AM, Mikko wrote:
On 2025-03-11 03:23:51 +0000, olcott said:
It may seem that way if you fail to understand
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
That you can quote some text but don't say anything about it
supports the
hypthesis that you don't understand the text you quoted.
*That you keep ignoring it is either stupid or dishonest*
What is that "it" intended to mean?
As a result, Y will stand forfoo(Y), which is
foo(foo(Y)) (because of what Y stands for),
which is foo(foo(foo(Y))), and so on. So Y ends
up standing for some kind of infinite structure.
Your quote is insufficient as the word "result" refers to someting
unquoted and does not mean anything without some context.
quoted from page 3
https://www.researchgate.net/ publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence
OK you are written off as a liar
Anyway, the quote has no relevance to anything I said in the previous
post.
On 3/12/2025 5:39 AM, Mikko wrote:
On 2025-03-11 23:10:14 +0000, olcott said:
On 3/11/2025 5:50 AM, Mikko wrote:
On 2025-03-11 03:23:51 +0000, olcott said:
It may seem that way if you fail to understand
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
That you can quote some text but don't say anything about it supports the >>>> hypthesis that you don't understand the text you quoted.
*That you keep ignoring it is either stupid or dishonest*
What is that "it" intended to mean?
As a result, Y will stand forfoo(Y), which is
foo(foo(Y)) (because of what Y stands for),
which is foo(foo(foo(Y))), and so on. So Y ends
up standing for some kind of infinite structure.
Your quote is insufficient as the word "result" refers to someting
unquoted and does not mean anything without some context.
quoted from page 3 https://www.researchgate.net/publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence
OK you are written off as a liar
Anyway, the quote has no relevance to anything I said in the previous post.
On 3/11/2025 5:50 AM, Mikko wrote:
On 2025-03-11 03:23:51 +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.
Right, the "Liar" is in the METALANGUAGE, not the LANGUAGE where
"It would then be possible to reconstruct the antinomy of the liar >>>>>>> in the metalanguage, by forming in the language itself a sentence" >>>>>>
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
It may seem that way if you fail to understand
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
That you can quote some text but don't say anything about it supports the
hypthesis that you don't understand the text you quoted.
I said that unify_with_occurs_check() detects
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".
On 3/15/2025 3:44 PM, Richard Damon wrote:
On 3/15/25 1:15 PM, olcott wrote:
On 3/11/2025 5:50 AM, Mikko wrote:
On 2025-03-11 03:23:51 +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
It may seem that way if you fail to understand
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
That you can quote some text but don't say anything about it
supports the
hypthesis that you don't understand the text you quoted.
I said that unify_with_occurs_check() detects
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".
Except for the fact that you aren't giving it the actual x that Tarski
creates (or the G for Godel) as expressed in the language, in part
because it uses logic that can't be expressed in Prolog.
Tarski's Liar Paradox from page 248
It would then be possible to reconstruct the antinomy of the liar
in the metalanguage, by forming in the language itself a sentence
x such that the sentence of the metalanguage which is correlated
with x asserts that x is not a true sentence.
https://liarparadox.org/Tarski_247_248.pdf
Formalized as:
x ∉ True if and only if p
where the symbol 'p' represents the whole sentence x https://liarparadox.org/Tarski_275_276.pdf
Not all all. It is merely that Tarski's somewhat clumsy
syntax does not encode the Liar Paradox where its
pathological self-reference can be directly seen.
He does not formalize most important part:
"where the symbol 'p' represents the whole sentence x"
If he did formalize that most important part it would
be this: x ∉ True if and only if x
On 3/15/2025 3:44 PM, Richard Damon wrote:
On 3/15/25 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.
Right, the "Liar" is in the METALANGUAGE, not the LANGUAGE where
"It would then be possible to reconstruct the antinomy of the liar >>>>>>> in the metalanguage, by forming in the language itself a sentence" >>>>>>
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.
WHen you claim that Prolog gives answers for logic system more
advanced then it, or make unsupported claims about your FRAUD of MTT,
you are just showing your stupidity.
Part of your problem, it seems, is that you don't understand the
limitations of Prolog, because you can't understand the logic that
Prolog can't handle, because you are just too stupid.
LP := ~True(LP
Try to explain in your own words what this means:
LP specifies a cycle in the directed graph of its evaluation sequence.
On 3/15/2025 9:12 PM, Richard Damon wrote:
On 3/15/25 9:19 PM, olcott wrote:
On 3/15/2025 3:44 PM, Richard Damon wrote:
On 3/15/25 1:15 PM, olcott wrote:
On 3/11/2025 5:50 AM, Mikko wrote:
On 2025-03-11 03:23:51 +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
It may seem that way if you fail to understand
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
That you can quote some text but don't say anything about it
supports the
hypthesis that you don't understand the text you quoted.
I said that unify_with_occurs_check() detects
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".
Except for the fact that you aren't giving it the actual x that
Tarski creates (or the G for Godel) as expressed in the language, in
part because it uses logic that can't be expressed in Prolog.
Tarski's Liar Paradox from page 248
It would then be possible to reconstruct the antinomy of the liar >>> in the metalanguage, by forming in the language itself a sentence >>> x such that the sentence of the metalanguage which is correlated
with x asserts that x is not a true sentence.
https://liarparadox.org/Tarski_247_248.pdf
Formalized as:
NO!!
That is what it reduces to in the metalangugae, but not what it is in
the language, which is where it counts.
x ∉ True if and only if p
where the symbol 'p' represents the whole sentence x
https://liarparadox.org/Tarski_275_276.pdf
Not all all. It is merely that Tarski's somewhat clumsy
syntax does not encode the Liar Paradox where its
pathological self-reference can be directly seen.
No, Tarski's syntax
He does not formalize most important part:
"where the symbol 'p' represents the whole sentence x"
If he did formalize that most important part it would
be this: x ∉ True if and only if x
Nope, you are just not understanding that 'x' is a fairly complecated
sentence in the language, for which in the metalanguge, it can be
reduced to the symbol p.
When Tarski formalized the Liar Paradox
HE DID IT INCORRECTLY.
LP := ~True(LP) <is> "This sentence is not true"
Tarski GOT THIS WRONG.
On 3/11/2025 5:50 AM, Mikko wrote:
On 2025-03-11 03:23:51 +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.
Right, the "Liar" is in the METALANGUAGE, not the LANGUAGE where the >>>>>> predicate is defined.
"It would then be possible to reconstruct the antinomy of the liar >>>>>>> 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 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
It may seem that way if you fail to understand
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
That you can quote some text but don't say anything about it supports the
hypthesis that you don't understand the text you quoted.
I said that unify_with_occurs_check() detects
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".
On 3/16/2025 6:33 AM, Richard Damon wrote:
On 3/15/25 10:37 PM, olcott wrote:reconstruct the antinomy of the liar
On 3/15/2025 9:12 PM, Richard Damon wrote:
On 3/15/25 9:19 PM, olcott wrote:
On 3/15/2025 3:44 PM, Richard Damon wrote:
On 3/15/25 1:15 PM, olcott wrote:
On 3/11/2025 5:50 AM, Mikko wrote:
On 2025-03-11 03:23:51 +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:
WHich is irrelevent, as that isn't the statement in view, >>>>>>>>>>>>>> only what could be shown to be a meaning of the actual >>>>>>>>>>>>>> statement.
LP := ~True(LP) DOES SPECIFY INFINITE RECURSION. >>>>>>>>>>>>>>
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
It may seem that way if you fail to understand
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
That you can quote some text but don't say anything about it
supports the
hypthesis that you don't understand the text you quoted.
I said that unify_with_occurs_check() detects
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".
Except for the fact that you aren't giving it the actual x that
Tarski creates (or the G for Godel) as expressed in the language,
in part because it uses logic that can't be expressed in Prolog.
Tarski's Liar Paradox from page 248
It would then be possible to reconstruct the antinomy of the liar >>>>> in the metalanguage, by forming in the language itself a sentence >>>>> x such that the sentence of the metalanguage which is correlated >>>>> with x asserts that x is not a true sentence.
https://liarparadox.org/Tarski_247_248.pdf
Formalized as:
NO!!
That is what it reduces to in the metalangugae, but not what it is
in the language, which is where it counts.
x ∉ True if and only if p
where the symbol 'p' represents the whole sentence x
https://liarparadox.org/Tarski_275_276.pdf
Not all all. It is merely that Tarski's somewhat clumsy
syntax does not encode the Liar Paradox where its
pathological self-reference can be directly seen.
No, Tarski's syntax
He does not formalize most important part:
"where the symbol 'p' represents the whole sentence x"
If he did formalize that most important part it would
be this: x ∉ True if and only if x
Nope, you are just not understanding that 'x' is a fairly
complecated sentence in the language, for which in the metalanguge,
it can be reduced to the symbol p.
When Tarski formalized the Liar Paradox
HE DID IT INCORRECTLY.
We wasn't "Formalizing" the Liar Paradox.
in the metalanguage, by forming in the language itself a sentence
x such that the sentence of the metalanguage which is correlated
with x asserts that x is not a true sentence.
LP := ~True(LP) <is> "This sentence is not true"
Tarski GOT THIS WRONG.
Nope, you don't understand what he is doing, because he is using
thought to get to a goal, something that seems to be beyond you.
You are just too stupid to understand the thoughts he is thinking
because you "logic" isn't correct, and too simple.
The issue is that you are a liar as I have shown above.
On 3/16/2025 5:50 PM, Richard Damon wrote:
On 3/16/25 10:32 AM, olcott wrote:
On 3/16/2025 6:33 AM, Richard Damon wrote:
On 3/15/25 10:37 PM, olcott wrote:reconstruct the antinomy of the liar
On 3/15/2025 9:12 PM, Richard Damon wrote:
On 3/15/25 9:19 PM, olcott wrote:
On 3/15/2025 3:44 PM, Richard Damon wrote:
On 3/15/25 1:15 PM, olcott wrote:
On 3/11/2025 5:50 AM, Mikko wrote:
On 2025-03-11 03:23:51 +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:
WHich is irrelevent, as that isn't the statement in >>>>>>>>>>>>>>>> view, only what could be shown to be a meaning of the >>>>>>>>>>>>>>>> actual statement.
LP := ~True(LP) DOES SPECIFY INFINITE RECURSION. >>>>>>>>>>>>>>>>
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
It may seem that way if you fail to understand
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
That you can quote some text but don't say anything about it >>>>>>>>>> supports the
hypthesis that you don't understand the text you quoted.
I said that unify_with_occurs_check() detects
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".
Except for the fact that you aren't giving it the actual x that >>>>>>>> Tarski creates (or the G for Godel) as expressed in the
language, in part because it uses logic that can't be expressed >>>>>>>> in Prolog.
Tarski's Liar Paradox from page 248
It would then be possible to reconstruct the antinomy of the >>>>>>> liar
in the metalanguage, by forming in the language itself a
sentence
x such that the sentence of the metalanguage which is correlated >>>>>>> with x asserts that x is not a true sentence.
https://liarparadox.org/Tarski_247_248.pdf
Formalized as:
NO!!
That is what it reduces to in the metalangugae, but not what it is >>>>>> in the language, which is where it counts.
x ∉ True if and only if p
where the symbol 'p' represents the whole sentence x
https://liarparadox.org/Tarski_275_276.pdf
Not all all. It is merely that Tarski's somewhat clumsy
syntax does not encode the Liar Paradox where its
pathological self-reference can be directly seen.
No, Tarski's syntax
He does not formalize most important part:
"where the symbol 'p' represents the whole sentence x"
If he did formalize that most important part it would
be this: x ∉ True if and only if x
Nope, you are just not understanding that 'x' is a fairly
complecated sentence in the language, for which in the
metalanguge, it can be reduced to the symbol p.
When Tarski formalized the Liar Paradox
HE DID IT INCORRECTLY.
We wasn't "Formalizing" the Liar Paradox.
in the metalanguage, by forming in the language itself a sentence >>> x such that the sentence of the metalanguage which is correlated >>> with x asserts that x is not a true sentence.
Apparently you don't understand what it means to "reconstruct" something.
Incorrectly reconstruct "the antinomy of the liar"
by some incorrect means. Tarski stupidly forgot to
include self-reference.
On 3/16/2025 8:19 AM, Mikko wrote:
On 2025-03-15 17:15:39 +0000, olcott said:
On 3/11/2025 5:50 AM, Mikko wrote:
On 2025-03-11 03:23:51 +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
It may seem that way if you fail to understand
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
That you can quote some text but don't say anything about it supports the >>>> hypthesis that you don't understand the text you quoted.
I said that unify_with_occurs_check() detects
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 above is irrelevant to the fact that you didn't say anothing about
the text you quoted.
LP := ~True(LP) expanded to infinite recursion ~True(~True(~True(~True(~True(~True(...))))))
The same way that Clocksin and Mellish do on their example
that you dishonestly keep ignoring.
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.
Right, the "Liar" is in the METALANGUAGE, not the LANGUAGE where the
"It would then be possible to reconstruct the antinomy of the liar
in the metalanguage, by forming in the language itself a sentence" >>>>
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.
On 3/16/2025 9:52 PM, Richard Damon wrote:
On 3/16/25 8:22 PM, olcott wrote:
On 3/16/2025 5:50 PM, Richard Damon wrote:
On 3/16/25 10:32 AM, olcott wrote:
On 3/16/2025 6:33 AM, Richard Damon wrote:
On 3/15/25 10:37 PM, olcott wrote:reconstruct the antinomy of the liar
On 3/15/2025 9:12 PM, Richard Damon wrote:
On 3/15/25 9:19 PM, olcott wrote:
On 3/15/2025 3:44 PM, Richard Damon wrote:
On 3/15/25 1:15 PM, olcott wrote:
On 3/11/2025 5:50 AM, Mikko wrote:
On 2025-03-11 03:23:51 +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:
WHich is irrelevent, as that isn't the statement in >>>>>>>>>>>>>>>>>> view, only what could be shown to be a meaning of the >>>>>>>>>>>>>>>>>> actual statement.
LP := ~True(LP) DOES SPECIFY INFINITE RECURSION. >>>>>>>>>>>>>>>>>>
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
It may seem that way if you fail to understand
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
That you can quote some text but don't say anything about it >>>>>>>>>>>> supports the
hypthesis that you don't understand the text you quoted. >>>>>>>>>>>>
I said that unify_with_occurs_check() detects
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".
Except for the fact that you aren't giving it the actual x >>>>>>>>>> that Tarski creates (or the G for Godel) as expressed in the >>>>>>>>>> language, in part because it uses logic that can't be
expressed in Prolog.
Tarski's Liar Paradox from page 248
It would then be possible to reconstruct the antinomy of >>>>>>>>> the liar
in the metalanguage, by forming in the language itself a >>>>>>>>> sentence
x such that the sentence of the metalanguage which is >>>>>>>>> correlated
with x asserts that x is not a true sentence.
https://liarparadox.org/Tarski_247_248.pdf
Formalized as:
NO!!
That is what it reduces to in the metalangugae, but not what it >>>>>>>> is in the language, which is where it counts.
x ∉ True if and only if p
where the symbol 'p' represents the whole sentence x
https://liarparadox.org/Tarski_275_276.pdf
Not all all. It is merely that Tarski's somewhat clumsy
syntax does not encode the Liar Paradox where its
pathological self-reference can be directly seen.
No, Tarski's syntax
He does not formalize most important part:
"where the symbol 'p' represents the whole sentence x"
If he did formalize that most important part it would
be this: x ∉ True if and only if x
Nope, you are just not understanding that 'x' is a fairly
complecated sentence in the language, for which in the
metalanguge, it can be reduced to the symbol p.
When Tarski formalized the Liar Paradox
HE DID IT INCORRECTLY.
We wasn't "Formalizing" the Liar Paradox.
in the metalanguage, by forming in the language itself a sentence
x such that the sentence of the metalanguage which is correlated >>>>> with x asserts that x is not a true sentence.
Apparently you don't understand what it means to "reconstruct"
something.
Incorrectly reconstruct "the antinomy of the liar"
by some incorrect means. Tarski stupidly forgot to
include self-reference.
Nope, you don't understand what he did.
Where is the error in the previous proof that shows how to construct
that x?
x ∉ True if and only if p
where the symbol 'p' represents the whole sentence x https://liarparadox.org/Tarski_275_276.pdf
That does not say: "This sentence is not true"
The self-reference is only in the English and not
encoded n the formalism thus cannot be directly
evaluated in the formalism.
This does say: LP := ~True(LP)
"This sentence is not true"
I guess you are just going to need to admit you are too stupid to
understand what he did.
He didn't "forget" to include self-reference, he just shows that you
can build references indirectly with mathematics in a way that it
isn't there in the logical form, Just like Godel did.
Perhaps the problem with Tarski is that his proof gets abstract enough
that it is harder to understand how he does it, but it is still done
correctly, unless you can point to an actual error.
As opposed to saying the answer doesn't match your OPINION of what it
should be, for which you have NO PROOF that your OPINION has any real
basis other than your admitted lies.
On 3/17/2025 6:26 AM, Richard Damon wrote:
On 3/17/25 12:36 AM, olcott wrote:
x ∉ True if and only if p
where the symbol 'p' represents the whole sentence x
https://liarparadox.org/Tarski_275_276.pdf
That does not say: "This sentence is not true"
The self-reference is only in the English and not
encoded n the formalism thus cannot be directly
evaluated in the formalism.
This does say: LP := ~True(LP)
"This sentence is not true"
But that sentence you started with is only in the METALANGUAGE, so
your "Formalism" isn't a statement in the LANGUAGE.
x is a fully defined expression in the language developed per that
earlier proof.
So, x doesn't NEED to be "formalized" as it IS formalized.
The issue is that the "self-reference" isn't anything expressed in the
LANGUAGE, so isn't part of x itself, but is based on properties
established in the METALANGUAGE that can be expressed in the language.
Sorry, you are just showing that you don't understand what you are
talking about.
There is no counter-example in the set of human general
knowledge that can be expressed using language such that
True(X) does not work correctly...
On 3/17/2025 6:26 AM, Richard Damon wrote:
On 3/17/25 12:36 AM, olcott wrote:
x ∉ True if and only if p
where the symbol 'p' represents the whole sentence x
https://liarparadox.org/Tarski_275_276.pdf
That does not say: "This sentence is not true"
The self-reference is only in the English and not
encoded n the formalism thus cannot be directly
evaluated in the formalism.
This does say: LP := ~True(LP)
"This sentence is not true"
But that sentence you started with is only in the METALANGUAGE, so your
"Formalism" isn't a statement in the LANGUAGE.
x is a fully defined expression in the language developed per that
earlier proof.
So, x doesn't NEED to be "formalized" as it IS formalized.
The issue is that the "self-reference" isn't anything expressed in the
LANGUAGE, so isn't part of x itself, but is based on properties
established in the METALANGUAGE that can be expressed in the language.
Sorry, you are just showing that you don't understand what you are
talking about.
There is no counter-example in the set of human general
knowledge that can be expressed using language such that
True(X) does not work correctly...
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.
Right, the "Liar" is in the METALANGUAGE, not the LANGUAGE where the >>>>>> predicate is defined.
"It would then be possible to reconstruct the antinomy of the liar >>>>>>> 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 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)).
?- unify_with_occurs_check(LP, not(true(LP))).
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)
On 3/18/2025 9:08 AM, Mikko wrote:
On 2025-03-17 15:56:38 +0000, olcott said:
On 3/17/2025 6:26 AM, Richard Damon wrote:
On 3/17/25 12:36 AM, olcott wrote:
x ∉ True if and only if p
where the symbol 'p' represents the whole sentence x
https://liarparadox.org/Tarski_275_276.pdf
That does not say: "This sentence is not true"
The self-reference is only in the English and not
encoded n the formalism thus cannot be directly
evaluated in the formalism.
This does say: LP := ~True(LP)
"This sentence is not true"
But that sentence you started with is only in the METALANGUAGE, so
your "Formalism" isn't a statement in the LANGUAGE.
x is a fully defined expression in the language developed per that
earlier proof.
So, x doesn't NEED to be "formalized" as it IS formalized.
The issue is that the "self-reference" isn't anything expressed in
the LANGUAGE, so isn't part of x itself, but is based on properties
established in the METALANGUAGE that can be expressed in the language. >>>>
Sorry, you are just showing that you don't understand what you are
talking about.
There is no counter-example in the set of human general
knowledge that can be expressed using language such that
True(X) does not work correctly...
That very much depends on what does "correctly" mean about "True(X)".
The common meaning of True on the basis of the meaning
of words such as "cats are animals" for all words
and all meanings.
On 3/18/2025 9:20 AM, Mikko wrote:
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.
I have been saying "cycles" all along and it has always been cycles.
https://www.swi-prolog.org/pldoc/man?predicate=unify_with_occurs_check/2
On 3/18/2025 10:04 PM, Richard Damon wrote:
On 3/18/25 9:36 PM, olcott wrote:
On 3/18/2025 9:08 AM, Mikko wrote:
On 2025-03-17 15:56:38 +0000, olcott said:
On 3/17/2025 6:26 AM, Richard Damon wrote:
On 3/17/25 12:36 AM, olcott wrote:
x ∉ True if and only if p
where the symbol 'p' represents the whole sentence x
https://liarparadox.org/Tarski_275_276.pdf
That does not say: "This sentence is not true"
The self-reference is only in the English and not
encoded n the formalism thus cannot be directly
evaluated in the formalism.
This does say: LP := ~True(LP)
"This sentence is not true"
But that sentence you started with is only in the METALANGUAGE, so >>>>>> your "Formalism" isn't a statement in the LANGUAGE.
x is a fully defined expression in the language developed per that >>>>>> earlier proof.
So, x doesn't NEED to be "formalized" as it IS formalized.
The issue is that the "self-reference" isn't anything expressed in >>>>>> the LANGUAGE, so isn't part of x itself, but is based on
properties established in the METALANGUAGE that can be expressed
in the language.
Sorry, you are just showing that you don't understand what you are >>>>>> talking about.
There is no counter-example in the set of human general
knowledge that can be expressed using language such that
True(X) does not work correctly...
That very much depends on what does "correctly" mean about "True(X)".
The common meaning of True on the basis of the meaning
of words such as "cats are animals" for all words
and all meanings.
And thus must include statements with infinite chains of inferences,
even when the results are unknown.
Sorry, your problem is you THINK you know what your words mean, but
don't actually understand all the implications, because you are just
too stupid.
I think that issue is actually your ADD because I have
qualified my claim for quite a while now limiting it
to THE SET OF HUMAN KNOWLEDGE THAT CAN BE EXPRESSED
USING LANGUAGE.
On 3/19/2025 10:57 AM, Mikko wrote:
On 2025-03-19 01:52:01 +0000, olcott said:
On 3/18/2025 9:20 AM, Mikko wrote:
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.
I have been saying "cycles" all along and it has always been cycles.
Not all along, just occasionally. What you did say that Prolog proves
that there is a cycle in the directed graph of their evaluation sequence.
I said that Prolog does not prove that. Then you posted some examples of
prolog not proving that and didn't mention "cycles" any more.
https://www.swi-prolog.org/pldoc/man?predicate=unify_with_occurs_check/2
That link confirms what I said above. It also said that one of the arguments >> already has a cycle then that cycle does not prevent unification and does
not cause infinite execution.
Clearly you have no idea what a cycle in a directed graph means.
On 3/18/2025 9:08 AM, Mikko wrote:
On 2025-03-17 15:56:38 +0000, olcott said:
On 3/17/2025 6:26 AM, Richard Damon wrote:
On 3/17/25 12:36 AM, olcott wrote:
x ∉ True if and only if p
where the symbol 'p' represents the whole sentence x
https://liarparadox.org/Tarski_275_276.pdf
That does not say: "This sentence is not true"
The self-reference is only in the English and not
encoded n the formalism thus cannot be directly
evaluated in the formalism.
This does say: LP := ~True(LP)
"This sentence is not true"
But that sentence you started with is only in the METALANGUAGE, so your >>>> "Formalism" isn't a statement in the LANGUAGE.
x is a fully defined expression in the language developed per that
earlier proof.
So, x doesn't NEED to be "formalized" as it IS formalized.
The issue is that the "self-reference" isn't anything expressed in the >>>> LANGUAGE, so isn't part of x itself, but is based on properties
established in the METALANGUAGE that can be expressed in the language. >>>>
Sorry, you are just showing that you don't understand what you are
talking about.
There is no counter-example in the set of human general
knowledge that can be expressed using language such that
True(X) does not work correctly...
That very much depends on what does "correctly" mean about "True(X)".
The common meaning of True on the basis of the meaning
of words such as "cats are animals" for all words
and all meanings.
On 3/17/2025 4:04 AM, Mikko wrote:
On 2025-03-16 14:38:16 +0000, olcott said:
On 3/16/2025 8:19 AM, Mikko wrote:
On 2025-03-15 17:15:39 +0000, olcott said:
On 3/11/2025 5:50 AM, Mikko wrote:
On 2025-03-11 03:23:51 +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
It may seem that way if you fail to understand
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
That you can quote some text but don't say anything about it supports the
hypthesis that you don't understand the text you quoted.
I said that unify_with_occurs_check() detects
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 above is irrelevant to the fact that you didn't say anothing about >>>> the text you quoted.
LP := ~True(LP) expanded to infinite recursion
~True(~True(~True(~True(~True(~True(...))))))
The same way that Clocksin and Mellish do on their example
that you dishonestly keep ignoring.
They don't say so in the above quoted text. What they do say is essentially >> what I have said in another context but not relevant here.
*It seems to me that you are dishonest abut that*
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)
On 3/20/2025 7:57 AM, Mikko wrote:
On 2025-03-17 13:18:42 +0000, olcott said:
On 3/17/2025 4:04 AM, Mikko wrote:
On 2025-03-16 14:38:16 +0000, olcott said:
On 3/16/2025 8:19 AM, Mikko wrote:
On 2025-03-15 17:15:39 +0000, olcott said:
On 3/11/2025 5:50 AM, Mikko wrote:
On 2025-03-11 03:23:51 +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:
WHich is irrelevent, as that isn't the statement in view, >>>>>>>>>>>>>> only what could be shown to be a meaning of the actual >>>>>>>>>>>>>> statement.
LP := ~True(LP) DOES SPECIFY INFINITE RECURSION. >>>>>>>>>>>>>>
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
It may seem that way if you fail to understand
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
That you can quote some text but don't say anything about it
supports the
hypthesis that you don't understand the text you quoted.
I said that unify_with_occurs_check() detects
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 above is irrelevant to the fact that you didn't say anothing
about
the text you quoted.
LP := ~True(LP) expanded to infinite recursion
~True(~True(~True(~True(~True(~True(...))))))
The same way that Clocksin and Mellish do on their example
that you dishonestly keep ignoring.
They don't say so in the above quoted text. What they do say is
essentially
what I have said in another context but not relevant here.
*It seems to me that you are dishonest abut that*
Doesn't matter. Hopefully readers can see that you are dishonest but
that is their problem, not yours or mine.
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 above quote is irrelevant to the question whether ~True(LP) resolves
to true.
If ?- equal(foo(Y), Y)
resolves to foo(foo(foo(foo(foo(foo(...))))))
then ?- LP = not(true(LP)).
resolves to not(true(not(true(not(true(not(true(...))))))))
On 3/20/2025 4:56 AM, Mikko wrote:
On 2025-03-19 01:36:33 +0000, olcott said:
On 3/18/2025 9:08 AM, Mikko wrote:
On 2025-03-17 15:56:38 +0000, olcott said:
On 3/17/2025 6:26 AM, Richard Damon wrote:
On 3/17/25 12:36 AM, olcott wrote:
x ∉ True if and only if p
where the symbol 'p' represents the whole sentence x
https://liarparadox.org/Tarski_275_276.pdf
That does not say: "This sentence is not true"
The self-reference is only in the English and not
encoded n the formalism thus cannot be directly
evaluated in the formalism.
This does say: LP := ~True(LP)
"This sentence is not true"
But that sentence you started with is only in the METALANGUAGE, so your >>>>>> "Formalism" isn't a statement in the LANGUAGE.
x is a fully defined expression in the language developed per that >>>>>> earlier proof.
So, x doesn't NEED to be "formalized" as it IS formalized.
The issue is that the "self-reference" isn't anything expressed in the >>>>>> LANGUAGE, so isn't part of x itself, but is based on properties
established in the METALANGUAGE that can be expressed in the language. >>>>>>
Sorry, you are just showing that you don't understand what you are >>>>>> talking about.
There is no counter-example in the set of human general
knowledge that can be expressed using language such that
True(X) does not work correctly...
That very much depends on what does "correctly" mean about "True(X)".
The common meaning of True on the basis of the meaning
of words such as "cats are animals" for all words
and all meanings.
Not possible (althogh a partial soultion could be useful).
The set of all human general knowledge that
can be expressed using language is complete.
On 3/20/2025 7:57 AM, Mikko wrote:
On 2025-03-17 13:18:42 +0000, olcott said:
On 3/17/2025 4:04 AM, Mikko wrote:
On 2025-03-16 14:38:16 +0000, olcott said:
On 3/16/2025 8:19 AM, Mikko wrote:
On 2025-03-15 17:15:39 +0000, olcott said:
On 3/11/2025 5:50 AM, Mikko wrote:
On 2025-03-11 03:23:51 +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:
WHich is irrelevent, as that isn't the statement in view, only what
LP := ~True(LP) DOES SPECIFY INFINITE RECURSION. >>>>>>>>>>>>>>
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
It may seem that way if you fail to understand
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
That you can quote some text but don't say anything about it supports the
hypthesis that you don't understand the text you quoted.
I said that unify_with_occurs_check() detects
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 above is irrelevant to the fact that you didn't say anothing about >>>>>> the text you quoted.
LP := ~True(LP) expanded to infinite recursion
~True(~True(~True(~True(~True(~True(...))))))
The same way that Clocksin and Mellish do on their example
that you dishonestly keep ignoring.
They don't say so in the above quoted text. What they do say is essentially
what I have said in another context but not relevant here.
*It seems to me that you are dishonest abut that*
Doesn't matter. Hopefully readers can see that you are dishonest but
that is their problem, not yours or mine.
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 above quote is irrelevant to the question whether ~True(LP) resolves
to true.
If ?- equal(foo(Y), Y)
resolves to foo(foo(foo(foo(foo(foo(...))))))
then ?- LP = not(true(LP)).
resolves to not(true(not(true(not(true(not(true(...))))))))
On 3/21/2025 4:00 AM, Mikko wrote:
On 2025-03-21 03:53:53 +0000, olcott said:The set of human general knowledge that can be expressed
On 3/20/2025 4:56 AM, Mikko wrote:
On 2025-03-19 01:36:33 +0000, olcott said:
On 3/18/2025 9:08 AM, Mikko wrote:
On 2025-03-17 15:56:38 +0000, olcott said:
On 3/17/2025 6:26 AM, Richard Damon wrote:
On 3/17/25 12:36 AM, olcott wrote:
x ∉ True if and only if p
where the symbol 'p' represents the whole sentence x
https://liarparadox.org/Tarski_275_276.pdf
That does not say: "This sentence is not true"
The self-reference is only in the English and not
encoded n the formalism thus cannot be directly
evaluated in the formalism.
This does say: LP := ~True(LP)
"This sentence is not true"
But that sentence you started with is only in the METALANGUAGE, >>>>>>>> so your "Formalism" isn't a statement in the LANGUAGE.
x is a fully defined expression in the language developed per
that earlier proof.
So, x doesn't NEED to be "formalized" as it IS formalized.
The issue is that the "self-reference" isn't anything expressed >>>>>>>> in the LANGUAGE, so isn't part of x itself, but is based on
properties established in the METALANGUAGE that can be expressed >>>>>>>> in the language.
Sorry, you are just showing that you don't understand what you >>>>>>>> are talking about.
There is no counter-example in the set of human general
knowledge that can be expressed using language such that
True(X) does not work correctly...
That very much depends on what does "correctly" mean about "True(X)". >>>>>>
The common meaning of True on the basis of the meaning
of words such as "cats are animals" for all words
and all meanings.
Not possible (althogh a partial soultion could be useful).
The set of all human general knowledge that
can be expressed using language is complete.
For some meaning of "complete".
A bigger problem is that it is infinite.
using language is finite and stored in a general knowledge
ontology. Specific situation and discourse context knowledge
ontologies can be defined as needed.
On 3/21/2025 4:00 AM, Mikko wrote:
On 2025-03-21 03:53:53 +0000, olcott said:The set of human general knowledge that can be expressed
On 3/20/2025 4:56 AM, Mikko wrote:
On 2025-03-19 01:36:33 +0000, olcott said:
On 3/18/2025 9:08 AM, Mikko wrote:
On 2025-03-17 15:56:38 +0000, olcott said:
On 3/17/2025 6:26 AM, Richard Damon wrote:
On 3/17/25 12:36 AM, olcott wrote:
x ∉ True if and only if p
where the symbol 'p' represents the whole sentence x
https://liarparadox.org/Tarski_275_276.pdf
That does not say: "This sentence is not true"
The self-reference is only in the English and not
encoded n the formalism thus cannot be directly
evaluated in the formalism.
This does say: LP := ~True(LP)
"This sentence is not true"
But that sentence you started with is only in the METALANGUAGE, so your
"Formalism" isn't a statement in the LANGUAGE.
x is a fully defined expression in the language developed per that >>>>>>>> earlier proof.
So, x doesn't NEED to be "formalized" as it IS formalized.
The issue is that the "self-reference" isn't anything expressed in the >>>>>>>> LANGUAGE, so isn't part of x itself, but is based on properties >>>>>>>> established in the METALANGUAGE that can be expressed in the language. >>>>>>>>
Sorry, you are just showing that you don't understand what you are >>>>>>>> talking about.
There is no counter-example in the set of human general
knowledge that can be expressed using language such that
True(X) does not work correctly...
That very much depends on what does "correctly" mean about "True(X)". >>>>>>
The common meaning of True on the basis of the meaning
of words such as "cats are animals" for all words
and all meanings.
Not possible (althogh a partial soultion could be useful).
The set of all human general knowledge that
can be expressed using language is complete.
For some meaning of "complete".
A bigger problem is that it is infinite.
using language is finite and stored in a general knowledge
ontology.
On 3/21/2025 4:06 AM, Mikko wrote:
On 2025-03-20 22:55:17 +0000, olcott said:
On 3/20/2025 7:57 AM, Mikko wrote:
On 2025-03-17 13:18:42 +0000, olcott said:
On 3/17/2025 4:04 AM, Mikko wrote:
On 2025-03-16 14:38:16 +0000, olcott said:
On 3/16/2025 8:19 AM, Mikko wrote:
On 2025-03-15 17:15:39 +0000, olcott said:
On 3/11/2025 5:50 AM, Mikko wrote:
On 2025-03-11 03:23:51 +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:
WHich is irrelevent, as that isn't the statement in view, only what
LP := ~True(LP) DOES SPECIFY INFINITE RECURSION. >>>>>>>>>>>>>>>>
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
It may seem that way if you fail to understand
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
That you can quote some text but don't say anything about it supports the
hypthesis that you don't understand the text you quoted.
I said that unify_with_occurs_check() detects
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 above is irrelevant to the fact that you didn't say anothing about >>>>>>>> the text you quoted.
LP := ~True(LP) expanded to infinite recursion
~True(~True(~True(~True(~True(~True(...))))))
The same way that Clocksin and Mellish do on their example
that you dishonestly keep ignoring.
They don't say so in the above quoted text. What they do say is essentially
what I have said in another context but not relevant here.
*It seems to me that you are dishonest abut that*
Doesn't matter. Hopefully readers can see that you are dishonest but
that is their problem, not yours or mine.
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 above quote is irrelevant to the question whether ~True(LP) resolves >>>> to true.
If ?- equal(foo(Y), Y)
resolves to foo(foo(foo(foo(foo(foo(...))))))
then ?- LP = not(true(LP)).
resolves to not(true(not(true(not(true(not(true(...))))))))
Of cours. But that is irrelevant to the fact that you quoted a text
without saying anything about it.
It is self-evident that both expressions specify infinite recursion.
You denied this so many times in so many ways it was as if you
formed your rebuttals without ever even glancing at any of my words.
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