LP := ~True(LP) is translated to Prolog:

?- 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)

https://www.researchgate.net/publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence --
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer

On 4/30/22 3:02 AM, olcott wrote:

LP := ~True(LP) is translated to Prolog:

?- 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)

https://www.researchgate.net/publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence

Since it isn't giving you a "syntax error", it is probably correct Prolog. Not sure if your interpretation of the results is correct.

All that false means is that the statement


LP = not(true(LP))

is recursive and that Prolog can't actually evaluate it due to its limited logic rules.

I will condition this answer on the fact that I am not a prolog specialist, but just reading the manual and providing basic understanding, which I am not sure of your ability to do so.

On 4/30/2022 8:08 PM, Richard Damon wrote:

On 4/30/22 3:02 AM, olcott wrote:

LP := ~True(LP) is translated to Prolog:

?- 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)

https://www.researchgate.net/publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence

Since it isn't giving you a "syntax error", it is probably correct Prolog. Not sure if your interpretation of the results is correct.

All that false means is that the statement


LP = not(true(LP))

is recursive and that Prolog can't actually evaluate it due to its limited logic rules.

That is not what Clocksin & Mellish says. They say it is an erroneous "infinite term" meaning that it specifies infinitely nested definition like this:

LP := ~True(LP) specifies:
~True(~True(~True(L~True(L~True(...))
The ellipses "..." mean "on and on forever"

One half a page of the Clocksin & Mellish text is quoted on page (3) of my paper:

https://www.researchgate.net/publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence

I will condition this answer on the fact that I am not a prolog specialist, but just reading the manual and providing basic understanding, which I am not sure of your ability to do so.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer

On 4/30/2022 8:08 PM, Richard Damon wrote:

On 4/30/22 3:02 AM, olcott wrote:

LP := ~True(LP) is translated to Prolog:

?- 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)

https://www.researchgate.net/publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence

Since it isn't giving you a "syntax error", it is probably correct Prolog. Not sure if your interpretation of the results is correct.

I asked the question incorrectly, what I really needed to know is whether or not the Prolog correctly encodes this logic sentence:
LP := ~True(LP)

x := y means x is defined to be another name for y
https://en.wikipedia.org/wiki/List_of_logic_symbols

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer

On 4/30/22 9:42 PM, olcott wrote:

On 4/30/2022 8:08 PM, Richard Damon wrote:

On 4/30/22 3:02 AM, olcott wrote:

LP := ~True(LP) is translated to Prolog:

?- 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)

https://www.researchgate.net/publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence

Since it isn't giving you a "syntax error", it is probably correct Prolog. Not sure if your interpretation of the results is correct.

All that false means is that the statement


LP = not(true(LP))

is recursive and that Prolog can't actually evaluate it due to its limited logic rules.

That is not what Clocksin & Mellish says. They say it is an erroneous "infinite term" meaning that it specifies infinitely nested definition like this:

No, that IS what they say, that this sort of recursion fails the test of Unification, not that it is has no possible logical meaning.

Prolog represents a somewhat basic form of logic, useful for many cases, but not encompassing all possible reasoning systems.

Maybe it can handle every one that YOU can understand, but it can't handle many higher order logical structures.

Note, for instance, at least some ways of writing factorial for an unknown value can lead to an infinite expansion, but the factorial is well defined for all positive integers. The fact that a "prolog like" expansion operator might not be able to handle the definition, doesn't mean it doesn't have meaning.

LP := ~True(LP) specifies:
~True(~True(~True(L~True(L~True(...))
The ellipses "..." mean "on and on forever"

One half a page of the Clocksin & Mellish text is quoted on page (3) of my paper:

https://www.researchgate.net/publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence

I will condition this answer on the fact that I am not a prolog specialist, but just reading the manual and providing basic understanding, which I am not sure of your ability to do so.

On 4/30/2022 9:00 PM, Richard Damon wrote:

On 4/30/22 9:42 PM, olcott wrote:

On 4/30/2022 8:08 PM, Richard Damon wrote:

On 4/30/22 3:02 AM, olcott wrote:

LP := ~True(LP) is translated to Prolog:

?- 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)

https://www.researchgate.net/publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence

Since it isn't giving you a "syntax error", it is probably correct Prolog. Not sure if your interpretation of the results is correct.

All that false means is that the statement


LP = not(true(LP))

is recursive and that Prolog can't actually evaluate it due to its limited logic rules.

That is not what Clocksin & Mellish says. They say it is an erroneous "infinite term" meaning that it specifies infinitely nested definition like this:

No, that IS what they say, that this sort of recursion fails the test of Unification, not that it is has no possible logical meaning.

Prolog represents a somewhat basic form of logic, useful for many cases, but not encompassing all possible reasoning systems.

Maybe it can handle every one that YOU can understand, but it can't handle many higher order logical structures.

Note, for instance, at least some ways of writing factorial for an unknown value can lead to an infinite expansion, but the factorial is well defined for all positive integers. The fact that a "prolog like" expansion operator might not be able to handle the definition, doesn't mean it doesn't have meaning.

It is really dumb that you continue to take wild guesses again the verified facts.

Please read the Clocksin & Mellish (on page 3 of my paper) text and eliminate your ignorance.

LP := ~True(LP) specifies:
~True(~True(~True(L~True(L~True(...))
The ellipses "..." mean "on and on forever"

One half a page of the Clocksin & Mellish text is quoted on page (3) of my paper:

https://www.researchgate.net/publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence

I will condition this answer on the fact that I am not a prolog specialist, but just reading the manual and providing basic understanding, which I am not sure of your ability to do so.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer

On 4/30/22 10:21 PM, olcott wrote:

On 4/30/2022 9:00 PM, Richard Damon wrote:

On 4/30/22 9:42 PM, olcott wrote:

On 4/30/2022 8:08 PM, Richard Damon wrote:

On 4/30/22 3:02 AM, olcott wrote:

LP := ~True(LP) is translated to Prolog:

?- 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)

https://www.researchgate.net/publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence

Since it isn't giving you a "syntax error", it is probably correct Prolog. Not sure if your interpretation of the results is correct.

All that false means is that the statement


LP = not(true(LP))

is recursive and that Prolog can't actually evaluate it due to its limited logic rules.

That is not what Clocksin & Mellish says. They say it is an erroneous "infinite term" meaning that it specifies infinitely nested definition like this:

No, that IS what they say, that this sort of recursion fails the test of Unification, not that it is has no possible logical meaning.

Prolog represents a somewhat basic form of logic, useful for many cases, but not encompassing all possible reasoning systems.

Maybe it can handle every one that YOU can understand, but it can't handle many higher order logical structures.

Note, for instance, at least some ways of writing factorial for an unknown value can lead to an infinite expansion, but the factorial is well defined for all positive integers. The fact that a "prolog like" expansion operator might not be able to handle the definition, doesn't mean it doesn't have meaning.

It is really dumb that you continue to take wild guesses again the verified facts.

Please read the Clocksin & Mellish (on page 3 of my paper) text and eliminate your ignorance.

I did. You just don't seem to understand what I am saying because it is above your head.

Prolog is NOT the defining authority for what is a valid logical statement, but a system of programming to handle a subset of those statements (a useful subset, but a subset).

The fact that Prolog doesn't allow something doesn't mean it doesn't have a logical meaning, only that it doesn't have a logical meaning in Prolog.

The inability of Prolog to "Unify" the expression, does not mean the expression doesn't have logical meaning, just that PROLOG can't derive meaning from the expression.

The Halting Problem and the incompleteness proofs never claims that they is designed for the subset of logic that is Prolog, and in fact they may be implicitly denying that, as I don't think Prolog handles enough complexity of logic to reach the threshold needed to express "G" in Godel's proof. (Your need to "simplify" it, is indicative of this, and shows you don't understand the actual proof).

This means that the fact that Prolog rejects unification of the statements doesn't actually say that much, just that the statement isn't of the type that Prolog can fully process.

On 4/30/2022 9:38 PM, Richard Damon wrote:

On 4/30/22 10:21 PM, olcott wrote:

On 4/30/2022 9:00 PM, Richard Damon wrote:

On 4/30/22 9:42 PM, olcott wrote:

On 4/30/2022 8:08 PM, Richard Damon wrote:

On 4/30/22 3:02 AM, olcott wrote:

LP := ~True(LP) is translated to Prolog:

?- 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)

https://www.researchgate.net/publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence

Since it isn't giving you a "syntax error", it is probably correct Prolog. Not sure if your interpretation of the results is correct.

All that false means is that the statement


LP = not(true(LP))

is recursive and that Prolog can't actually evaluate it due to its limited logic rules.

That is not what Clocksin & Mellish says. They say it is an erroneous "infinite term" meaning that it specifies infinitely nested definition like this:

No, that IS what they say, that this sort of recursion fails the test of Unification, not that it is has no possible logical meaning.

Prolog represents a somewhat basic form of logic, useful for many cases, but not encompassing all possible reasoning systems.

Maybe it can handle every one that YOU can understand, but it can't handle many higher order logical structures.

Note, for instance, at least some ways of writing factorial for an unknown value can lead to an infinite expansion, but the factorial is well defined for all positive integers. The fact that a "prolog like" expansion operator might not be able to handle the definition, doesn't mean it doesn't have meaning.

It is really dumb that you continue to take wild guesses again the verified facts.

Please read the Clocksin & Mellish (on page 3 of my paper) text and eliminate your ignorance.

I did. You just don't seem to understand what I am saying because it is above your head.

Prolog is NOT the defining authority for what is a valid logical statement, but a system of programming to handle a subset of those statements (a useful subset, but a subset).

The fact that Prolog doesn't allow something doesn't mean it doesn't have a logical meaning, only that it doesn't have a logical meaning in Prolog.

In this case it does. I have spent thousands of hours on the semantic error of infinitely recursive definition and written a dozen papers on it. Glancing at one of two of the words of Clocksin & Mellish does not count as reading it.

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 so on. So Y ends up standing for some kind of infinite structure.
END:(Clocksin & Mellish 2003:254)

foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(...)))))))))))))))


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer

On 4/30/22 10:56 PM, olcott wrote:

On 4/30/2022 9:38 PM, Richard Damon wrote:

On 4/30/22 10:21 PM, olcott wrote:

On 4/30/2022 9:00 PM, Richard Damon wrote:

On 4/30/22 9:42 PM, olcott wrote:

On 4/30/2022 8:08 PM, Richard Damon wrote:

On 4/30/22 3:02 AM, olcott wrote:

LP := ~True(LP) is translated to Prolog:

?- 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)

https://www.researchgate.net/publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence

Since it isn't giving you a "syntax error", it is probably correct Prolog. Not sure if your interpretation of the results is correct.

All that false means is that the statement


LP = not(true(LP))

is recursive and that Prolog can't actually evaluate it due to its limited logic rules.

That is not what Clocksin & Mellish says. They say it is an erroneous "infinite term" meaning that it specifies infinitely nested definition like this:

No, that IS what they say, that this sort of recursion fails the test of Unification, not that it is has no possible logical meaning.

Prolog represents a somewhat basic form of logic, useful for many cases, but not encompassing all possible reasoning systems.

Maybe it can handle every one that YOU can understand, but it can't handle many higher order logical structures.

Note, for instance, at least some ways of writing factorial for an unknown value can lead to an infinite expansion, but the factorial is well defined for all positive integers. The fact that a "prolog like" expansion operator might not be able to handle the definition, doesn't mean it doesn't have meaning.

It is really dumb that you continue to take wild guesses again the verified facts.

Please read the Clocksin & Mellish (on page 3 of my paper) text and eliminate your ignorance.

I did. You just don't seem to understand what I am saying because it is above your head.

Prolog is NOT the defining authority for what is a valid logical statement, but a system of programming to handle a subset of those statements (a useful subset, but a subset).

The fact that Prolog doesn't allow something doesn't mean it doesn't have a logical meaning, only that it doesn't have a logical meaning in Prolog.

In this case it does. I have spent thousands of hours on the semantic error of infinitely recursive definition and written a dozen papers on it. Glancing at one of two of the words of Clocksin & Mellish does not count as reading it.

And it appears that you don't understand it, because you still make category errors when trying to talk about it.

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 so on. So Y ends up standing for some kind of infinite structure.
END:(Clocksin & Mellish 2003:254)

foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(...)))))))))))))))

Right. but some infinite structures might actually have meaning. The fact that Prolog uses certain limited method to figure out meaning doesn't mean that other methods can't find the meaning.

Just like:

Fact(n) := (N == 1) ? 1 : N*Fact(n-1);

if naively expanded has an infinite expansion.

But, based on mathematical knowledge, and can actually be proven from the definition, something like Fact(n+1)/fact(n), even for an unknown n, can be reduced without the need to actually expend infinite operations.

Note, this is actual shown in your case of H(H^,H^). Yes, if H doesn't abort its simulation, then for THAT H^, we have that H^(H^) is non-halting, but so is H(H^,H^), and thus THAT H / H^ pair fails to be a counter example

When you program H to abort its simulation of H^ at some point, and build your H^ on that H, then H(H^,H^), will return the non-halting answer, and H^(H^) when PROPERLY run or simulated halts, because H has the same "cut off" logic at the factorial above.

The naive expansion thinks it is infinite, but the correct expansion sees the cut off and sees that it is actually finite.

On 4/30/2022 10:11 PM, Richard Damon wrote:

On 4/30/22 10:56 PM, olcott wrote:

On 4/30/2022 9:38 PM, Richard Damon wrote:

On 4/30/22 10:21 PM, olcott wrote:

On 4/30/2022 9:00 PM, Richard Damon wrote:

On 4/30/22 9:42 PM, olcott wrote:

On 4/30/2022 8:08 PM, Richard Damon wrote:

On 4/30/22 3:02 AM, olcott wrote:

LP := ~True(LP) is translated to Prolog:

?- 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)

https://www.researchgate.net/publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence

Since it isn't giving you a "syntax error", it is probably correct Prolog. Not sure if your interpretation of the results is correct.

All that false means is that the statement


LP = not(true(LP))

is recursive and that Prolog can't actually evaluate it due to its limited logic rules.

That is not what Clocksin & Mellish says. They say it is an erroneous "infinite term" meaning that it specifies infinitely nested definition like this:

No, that IS what they say, that this sort of recursion fails the test of Unification, not that it is has no possible logical meaning.

Prolog represents a somewhat basic form of logic, useful for many cases, but not encompassing all possible reasoning systems.

Maybe it can handle every one that YOU can understand, but it can't handle many higher order logical structures.

Note, for instance, at least some ways of writing factorial for an unknown value can lead to an infinite expansion, but the factorial is well defined for all positive integers. The fact that a "prolog like" expansion operator might not be able to handle the definition, doesn't mean it doesn't have meaning.

It is really dumb that you continue to take wild guesses again the verified facts.

Please read the Clocksin & Mellish (on page 3 of my paper) text and eliminate your ignorance.

I did. You just don't seem to understand what I am saying because it is above your head.

Prolog is NOT the defining authority for what is a valid logical statement, but a system of programming to handle a subset of those statements (a useful subset, but a subset).

The fact that Prolog doesn't allow something doesn't mean it doesn't have a logical meaning, only that it doesn't have a logical meaning in Prolog.

In this case it does. I have spent thousands of hours on the semantic error of infinitely recursive definition and written a dozen papers on it. Glancing at one of two of the words of Clocksin & Mellish does not count as reading it.

And it appears that you don't understand it, because you still make category errors when trying to talk about it.

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 so on. So Y ends up standing for some kind of infinite structure.
END:(Clocksin & Mellish 2003:254)

foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(...)))))))))))))))

Right. but some infinite structures might actually have meaning.

Not in this case, it is very obvious that no theorem prover can possibly prove any infinite expression. It is the same thing as a program that is stuck in an infinite loop.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer

On 4/30/2022 8:53 PM, André G. Isaak wrote:

On 2022-04-30 19:47, olcott wrote:

On 4/30/2022 8:08 PM, Richard Damon wrote:

On 4/30/22 3:02 AM, olcott wrote:

LP := ~True(LP) is translated to Prolog:

?- 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)

https://www.researchgate.net/publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence

Since it isn't giving you a "syntax error", it is probably correct Prolog. Not sure if your interpretation of the results is correct.

I asked the question incorrectly, what I really needed to know is whether or not the Prolog correctly encodes this logic sentence:
LP := ~True(LP)

Since that isn't a 'logic sentence', no one can answer this.

André

What about this one?
LP ↔ ¬True(LP)  // Tarski uses something like this
https://liarparadox.org/Tarski_275_276.pdf

or this one?
G ↔ ¬(F ⊢ G)



--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer

On 4/30/2022 9:15 PM, olcott wrote:

On 4/30/2022 10:11 PM, Richard Damon wrote:

On 4/30/22 10:56 PM, olcott wrote:

On 4/30/2022 9:38 PM, Richard Damon wrote:

On 4/30/22 10:21 PM, olcott wrote:

On 4/30/2022 9:00 PM, Richard Damon wrote:

On 4/30/22 9:42 PM, olcott wrote:

On 4/30/2022 8:08 PM, Richard Damon wrote:

On 4/30/22 3:02 AM, olcott wrote:

LP := ~True(LP) is translated to Prolog:

?- 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)

https://www.researchgate.net/publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence

Since it isn't giving you a "syntax error", it is probably
correct Prolog. Not sure if your interpretation of the results
is correct.

All that false means is that the statement


LP = not(true(LP))

is recursive and that Prolog can't actually evaluate it due to
its limited logic rules.

That is not what Clocksin & Mellish says. They say it is an
erroneous "infinite term" meaning that it specifies infinitely
nested definition like this:

No, that IS what they say, that this sort of recursion fails the
test of Unification, not that it is has no possible logical meaning.

Prolog represents a somewhat basic form of logic, useful for many
cases, but not encompassing all possible reasoning systems.

Maybe it can handle every one that YOU can understand, but it
can't handle many higher order logical structures.

Note, for instance, at least some ways of writing factorial for an
unknown value can lead to an infinite expansion, but the factorial
is well defined for all positive integers. The fact that a "prolog
like" expansion operator might not be able to handle the
definition, doesn't mean it doesn't have meaning.

It is really dumb that you continue to take wild guesses again the
verified facts.

Please read the Clocksin & Mellish (on page 3 of my paper) text and
eliminate your ignorance.

I did. You just don't seem to understand what I am saying because it
is above your head.

Prolog is NOT the defining authority for what is a valid logical
statement, but a system of programming to handle a subset of those
statements (a useful subset, but a subset).

The fact that Prolog doesn't allow something doesn't mean it doesn't
have a logical meaning, only that it doesn't have a logical meaning
in Prolog.

In this case it does. I have spent thousands of hours on the semantic
error of infinitely recursive definition and written a dozen papers
on it. Glancing at one of two of the words of Clocksin & Mellish does
not count as reading it.

And it appears that you don't understand it, because you still make
category errors when trying to talk about it.

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 so on. So Y ends up standing for some kind of
infinite structure.
END:(Clocksin & Mellish 2003:254)

foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(...)))))))))))))))

Right. but some infinite structures might actually have meaning.

Not in this case, it is very obvious that no theorem prover can possibly
prove any infinite expression. It is the same thing as a program that is
stuck in an infinite loop.

Richard wrote and the asshole (PO) snipped
------------------------------------------
Right. but some infinite structures might actually have meaning. The
fact that Prolog uses certain limited method to figure out meaning
doesn't mean that other methods can't find the meaning.

Just like:

Fact(n) := (N == 1) ? 1 : N*Fact(n-1);

if naively expanded has an infinite expansion.

But, based on mathematical knowledge, and can actually be proven from
the definition, something like Fact(n+1)/fact(n), even for an unknown n,
can be reduced without the need to actually expend infinite operations.

Note, this is actual shown in your case of H(H^,H^). Yes, if H doesn't
abort its simulation, then for THAT H^, we have that H^(H^) is
non-halting, but so is H(H^,H^), and thus THAT H / H^ pair fails to be a
counter example

When you program H to abort its simulation of H^ at some point, and
build your H^ on that H, then H(H^,H^), will return the non-halting
answer, and H^(H^) when PROPERLY run or simulated halts, because H has
the same "cut off" logic at the factorial above.

The naive expansion thinks it is infinite, but the correct expansion
sees the cut off and sees that it is actually finite.
----------------------------------------------------------------------

A good symbolic manipulation system or a theorem prover with appropriate
axioms and rules of inference could surely handle forms such as
Fact(n+1)/fact(n) without breathing hard. It is only you, an ignorant
fool, who seems to think that the unthinking infinite unrolling of a
form must occur. Only you would think that a solver system would
completely unroll a form before analyzing it and applying
transformations to it.

Son, it don't work that way (unless you are defining and making a mess
trying to write the system yourself). Systems usually have rules that
make small incremental transformations and usually search breadth first
with perhaps a limited amount of depth first interludes. If they don't
use a breadth first strategy, they will not be able to claim the
completeness property. (See resolution theorem prover literature for
some explanation. You wont understand it but you can cite as if you did!)

Richard was trying to explain this to you in the snipped portion I
recited just above. Question for Peter holding his pecker: How do you
always and I mean always manage to delete the part of a message you
respond too that addresses the point you now try to make?

A typically subsequence you might see in the trace: would include in
order but not necessarily consecutively:
    Fact(n+1)/Fact(n)
    (n+1)*Fact(n)/Fact(n)
    (n+1)
Some interspersed terms such as Fact(n+1)/(n*Fact(n-1)) would be found
too. In some circumstances, these other terms might be helpful. A
theorem prover or manipulator does all of this, breadth first, hoping to
blindly stumble on a solution. You can provide heuristics that might
speed up the process but no advice short of an oracle will get you even
one more result. (Another manifestation of HP.) It's the slow grinding
through the possibilities that guarantees that if a result can be found,
it will be found. And all the theory that you don't understand says
that's the best you can do.

Ben and I disagree on reasons for your type of total dishonesty. He
thinks that you are so self deluded that you actual believe what you are
saying; that you are so self deluded that the dishonest utterances are
just your subconscious protecting your already damaged ego. To that, I
say phooey; you are just a long term troll who lies a lot about math,
about your history and health, and about your accomplishments.

I don't believe that you will read this before you start to respond but
that's okay. Understanding is not required. Neither is respect in
either direction.
--
Jeff Barnett

On 5/1/2022 4:24 AM, Mikko wrote:

On 2022-04-30 18:15:19 +0000, Aleksy Grabowski said:

I just want to add some small note.

This "not(true(_))" thing is misleading.
Please note that it is *not* a negation.
Functionally it is equivalent to:

X = foo(bar(X)).

That's correct. Prolog language does not give any inherent semantics to
data structures. It only defines the execution semantics of language
structures and standard library symbols. Those same synbols can be used
in data structures with entirely different purposes.

Mikko

negation, not, \+
The concept of logical negation in Prolog is problematical, in the sense that the only method that Prolog can use to tell if a proposition is false is to try to prove it (from the facts and rules that it has been told about), and then if this attempt fails, it concludes that the proposition is false. This is referred to as negation as failure.

http://www.cse.unsw.edu.au/~billw/dictionaries/prolog/negation.html
This is actually a superior model to convention logic in that it only seeks to prove not true, thus detects expressions of language that are simply not truth bearers.


Expressions of (formal or natural) language that can possibly be resolved to a truth value are [truth bearers].

There are only two ways that an expression of language can be resolved to a truth value:
(1) An expression of language is assigned a truth value such as "cats are animals" is defined to be true.

(2) Truth preserving operations are applied to expressions of language that are known to be true. {cats are animals} and {animals are living things} therefore {cats are living things}. Copyright 2021 PL Olcott


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer

On 5/1/2022 4:26 AM, Mikko wrote:

On 2022-04-30 21:08:05 +0000, olcott said:

negation, not, \+
The concept of logical negation in Prolog is problematical, in the sense that the only method that Prolog can use to tell if a proposition is false is to try to prove it (from the facts and rules that it has been told about), and then if this attempt fails, it concludes that the proposition is false. This is referred to as negation as failure.

Note that the negation discussed above is not present in LP = not(true(LP)).

Mikko

Is says that it is. It says that "not" is synonymous with \+.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer

On 5/1/2022 4:38 AM, Mikko wrote:

On 2022-04-30 20:48:47 +0000, olcott said:

On 4/30/2022 4:31 AM, Mikko wrote:

On 2022-04-30 07:02:23 +0000, olcott said:

LP := ~True(LP) is translated to Prolog:

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

This is correct but to fail would also be correct.

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

unify_with_occurs_check must fail if the unified data structure
would contain loops.

Mikko

The above is the actual execution of actual Prolog code using
(SWI-Prolog (threaded, 64 bits, version 7.6.4).

Another Prolog implementation might interprete LP = not(true(LP)) differently
and still conform to the prolog standard.

According to Clocksin & Mellish it is not a mere loop, it is an "infinite term" thus infinitely recursive definition.

When discussing data structures, "infinite" and "loop" mean the same.
The data structure is infinitely deep but contains only finitely many
distinct objects and occupies only a finite amount of memory.

That is incorrect. any structure that is infinitely deep would take all of the memory that is available yet specifies an infinite amount of memory.

I am trying to validate whether or not my Prolog code encodes the Liar Paradox.

That cannot be inferred from Prolog rules. Prolog defines some encodings
like how to encode numbers with characters of Prolog character set but for
more complex things you must make your own encoding rules.

Mikko

This says that G is logically equivalent to its own unprovability in F
G ↔ ¬(F ⊢ G) and fails unify_with_occurs_check when encoded in Prolog.


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer

On 5/1/22 6:58 AM, olcott wrote:

On 5/1/2022 4:24 AM, Mikko wrote:

On 2022-04-30 18:15:19 +0000, Aleksy Grabowski said:

I just want to add some small note.

This "not(true(_))" thing is misleading.
Please note that it is *not* a negation.
Functionally it is equivalent to:

X = foo(bar(X)).

That's correct. Prolog language does not give any inherent semantics to
data structures. It only defines the execution semantics of language
structures and standard library symbols. Those same synbols can be used
in data structures with entirely different purposes.

Mikko

negation, not, \+
The concept of logical negation in Prolog is problematical, in the sense that the only method that Prolog can use to tell if a proposition is false is to try to prove it (from the facts and rules that it has been told about), and then if this attempt fails, it concludes that the proposition is false. This is referred to as negation as failure.

http://www.cse.unsw.edu.au/~billw/dictionaries/prolog/negation.html
This is actually a superior model to convention logic in that it only seeks to prove not true, thus detects expressions of language that are simply not truth bearers.


Expressions of (formal or natural) language that can possibly be resolved to a truth value are [truth bearers].

There are only two ways that an expression of language can be resolved to a truth value:
(1) An expression of language is assigned a truth value such as "cats are animals" is defined to be true.

(2) Truth preserving operations are applied to expressions of language that are known to be true. {cats are animals} and {animals are living things} therefore {cats are living things}. Copyright 2021 PL Olcott

So you are sort of answering your own question. The model of logic that Prolog handles isn't quite the same as "conventional" logic, in part due to the way it (doesn't) define Logical Negation.

This seems to fit into your standard misunderstanding of things.

On 4/30/22 11:15 PM, olcott wrote:

On 4/30/2022 10:11 PM, Richard Damon wrote:

On 4/30/22 10:56 PM, olcott wrote:

On 4/30/2022 9:38 PM, Richard Damon wrote:

On 4/30/22 10:21 PM, olcott wrote:

On 4/30/2022 9:00 PM, Richard Damon wrote:

On 4/30/22 9:42 PM, olcott wrote:

On 4/30/2022 8:08 PM, Richard Damon wrote:

On 4/30/22 3:02 AM, olcott wrote:

LP := ~True(LP) is translated to Prolog:

?- 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)

https://www.researchgate.net/publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence

Since it isn't giving you a "syntax error", it is probably correct Prolog. Not sure if your interpretation of the results is correct.

All that false means is that the statement


LP = not(true(LP))

is recursive and that Prolog can't actually evaluate it due to its limited logic rules.

That is not what Clocksin & Mellish says. They say it is an erroneous "infinite term" meaning that it specifies infinitely nested definition like this:

No, that IS what they say, that this sort of recursion fails the test of Unification, not that it is has no possible logical meaning.

Prolog represents a somewhat basic form of logic, useful for many cases, but not encompassing all possible reasoning systems.

Maybe it can handle every one that YOU can understand, but it can't handle many higher order logical structures.

Note, for instance, at least some ways of writing factorial for an unknown value can lead to an infinite expansion, but the factorial is well defined for all positive integers. The fact that a "prolog like" expansion operator might not be able to handle the definition, doesn't mean it doesn't have meaning.

It is really dumb that you continue to take wild guesses again the verified facts.

Please read the Clocksin & Mellish (on page 3 of my paper) text and eliminate your ignorance.

I did. You just don't seem to understand what I am saying because it is above your head.

Prolog is NOT the defining authority for what is a valid logical statement, but a system of programming to handle a subset of those statements (a useful subset, but a subset).

The fact that Prolog doesn't allow something doesn't mean it doesn't have a logical meaning, only that it doesn't have a logical meaning in Prolog.

In this case it does. I have spent thousands of hours on the semantic error of infinitely recursive definition and written a dozen papers on it. Glancing at one of two of the words of Clocksin & Mellish does not count as reading it.

And it appears that you don't understand it, because you still make category errors when trying to talk about it.

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 so on. So Y ends up standing for some kind of infinite structure.
END:(Clocksin & Mellish 2003:254)

foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(...)))))))))))))))

Right. but some infinite structures might actually have meaning.

Not in this case, it is very obvious that no theorem prover can possibly prove any infinite expression. It is the same thing as a program that is stuck in an infinite loop.

As Jeff pointed out, your claim is shown false, that some statements with infinite expansions can be worked with by some automatic solvers. This just proves that you don't really understand the effects of "recursion" and "self-reference", and perhaps a source of your error in trying to reason about thing like this Halting Problem proof or Godels Incompleteness Theory and his expression "G".

If something that is obviously undoable is your mind is shown to be in fact doable, the problem isn't in the ability to do it, but in the ability of your mind to understand. Your misconceptions don't change the nature of reality, but reality proves you wrong.

On 5/1/22 7:06 AM, olcott wrote:

On 5/1/2022 4:38 AM, Mikko wrote:

On 2022-04-30 20:48:47 +0000, olcott said:

On 4/30/2022 4:31 AM, Mikko wrote:

On 2022-04-30 07:02:23 +0000, olcott said:

LP := ~True(LP) is translated to Prolog:

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

This is correct but to fail would also be correct.

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

unify_with_occurs_check must fail if the unified data structure
would contain loops.

Mikko

The above is the actual execution of actual Prolog code using
(SWI-Prolog (threaded, 64 bits, version 7.6.4).

Another Prolog implementation might interprete LP = not(true(LP)) differently
and still conform to the prolog standard.

According to Clocksin & Mellish it is not a mere loop, it is an "infinite term" thus infinitely recursive definition.

When discussing data structures, "infinite" and "loop" mean the same.
The data structure is infinitely deep but contains only finitely many
distinct objects and occupies only a finite amount of memory.

That is incorrect. any structure that is infinitely deep would take all of the memory that is available yet specifies an infinite amount of memory.

Nope, a tree that one branch points into itself higher up represents a tree with infinite depth, but only needs a finite amount of memory. Building such a structure may require the ability to forward declare something or reference something not yet defined.

Some infinities have finite representation. You don't seem able to understand that.

Yes, some naive ways of expanding them fail, but the answer to that is you just don't do that, but need to use a less naive method.

I am trying to validate whether or not my Prolog code encodes the Liar Paradox.

That cannot be inferred from Prolog rules. Prolog defines some encodings
like how to encode numbers with characters of Prolog character set but for
more complex things you must make your own encoding rules.

Mikko

This says that G is logically equivalent to its own unprovability in F
G ↔ ¬(F ⊢ G) and fails unify_with_occurs_check when encoded in Prolog.

On 5/1/2022 4:45 AM, Mikko wrote:

On 2022-05-01 03:15:56 +0000, olcott said:

Not in this case, it is very obvious that no theorem prover can possibly prove any infinite expression.

Doesn't matter as you can't give an infinite expression to a theorem
prover.

Mikko

*You can and Prolog can detect and reject it*

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 so on. So Y ends up standing for some kind of infinite structure.

<inserted for clarity>
foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(...))))))))))))
</inserted for clarity>

Note that, whereas they may allow you to construct something like this, most Prolog systems will not be able to write it out at the end. According to the formal definition of Unification, this kind of “infinite term” should never come to exist. Thus Prolog systems that allow a term to match an uninstantiated subterm of itself do not act correctly as Resolution theorem provers. In order to make them do so, we would have to add a check that a variable cannot be instantiated to something containing itself. Such a check, an occurs check, would be straightforward to implement, but would slow down the execution of Prolog programs considerably. Since it would only affect very few programs, most implementors have simply left it out 1.

1 The Prolog standard states that the result is undefined if a Prolog system attempts to match a term against an uninstantiated subterm of itself, which means that programs which cause this  to happen will not be portable. A portable program should ensure that wherever an occurs check might be applicable the built-in predicate unify_with_occurs_check/2 is used explicitly instead of the normal unification operation of the Prolog implementation. As its name suggests, this predicate acts like =/2 except that it fails if an occurs check detects an illegal attempt to instantiate a variable.
END:(Clocksin & Mellish 2003:254)

Clocksin, W.F. and Mellish, C.S. 2003. Programming in Prolog Using the ISO Standard Fifth Edition, 254. Berlin Heidelberg: Springer-Verlag.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer

On 5/1/2022 12:24 AM, Jeff Barnett wrote:

On 4/30/2022 9:15 PM, olcott wrote:

On 4/30/2022 10:11 PM, Richard Damon wrote:

On 4/30/22 10:56 PM, olcott wrote:

On 4/30/2022 9:38 PM, Richard Damon wrote:

On 4/30/22 10:21 PM, olcott wrote:

On 4/30/2022 9:00 PM, Richard Damon wrote:

On 4/30/22 9:42 PM, olcott wrote:

On 4/30/2022 8:08 PM, Richard Damon wrote:

On 4/30/22 3:02 AM, olcott wrote:

LP := ~True(LP) is translated to Prolog:

?- 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)

https://www.researchgate.net/publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence

Since it isn't giving you a "syntax error", it is probably correct Prolog. Not sure if your interpretation of the results is correct.

All that false means is that the statement


LP = not(true(LP))

is recursive and that Prolog can't actually evaluate it due to its limited logic rules.

That is not what Clocksin & Mellish says. They say it is an erroneous "infinite term" meaning that it specifies infinitely nested definition like this:

No, that IS what they say, that this sort of recursion fails the test of Unification, not that it is has no possible logical meaning.

Prolog represents a somewhat basic form of logic, useful for many cases, but not encompassing all possible reasoning systems.

Maybe it can handle every one that YOU can understand, but it can't handle many higher order logical structures.

Note, for instance, at least some ways of writing factorial for an unknown value can lead to an infinite expansion, but the factorial is well defined for all positive integers. The fact that a "prolog like" expansion operator might not be able to handle the definition, doesn't mean it doesn't have meaning.

It is really dumb that you continue to take wild guesses again the verified facts.

Please read the Clocksin & Mellish (on page 3 of my paper) text and eliminate your ignorance.

I did. You just don't seem to understand what I am saying because it is above your head.

Prolog is NOT the defining authority for what is a valid logical statement, but a system of programming to handle a subset of those statements (a useful subset, but a subset).

The fact that Prolog doesn't allow something doesn't mean it doesn't have a logical meaning, only that it doesn't have a logical meaning in Prolog.

In this case it does. I have spent thousands of hours on the semantic error of infinitely recursive definition and written a dozen papers on it. Glancing at one of two of the words of Clocksin & Mellish does not count as reading it.

And it appears that you don't understand it, because you still make category errors when trying to talk about it.

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 so on. So Y ends up standing for some kind of infinite structure.
END:(Clocksin & Mellish 2003:254)

foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(...)))))))))))))))

Right. but some infinite structures might actually have meaning.

Not in this case, it is very obvious that no theorem prover can possibly prove any infinite expression. It is the same thing as a program that is stuck in an infinite loop.

Richard wrote and the asshole (PO) snipped
------------------------------------------
Right. but some infinite structures might actually have meaning. The fact that Prolog uses certain limited method to figure out meaning doesn't mean that other methods can't find the meaning.

The question is not whether some infinite structures have meaning that is the dishonest dodge of the strawman error.

The question is whether on not the expression at hand has meaning or is simply semantically incoherent. I just posted all of the Clocksin & Mellish text in my prior post to make this more clear.


This example expanded from Clocksin & Mellish conclusively proves that some expressions of language are incorrect:

foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(...))))))))))))

Just like:

Fact(n) := (N == 1) ? 1 : N*Fact(n-1);

if naively expanded has an infinite expansion.

But, based on mathematical knowledge, and can actually be proven from the definition, something like Fact(n+1)/fact(n), even for an unknown n, can be reduced without the need to actually expend infinite operations.

Note, this is actual shown in your case of H(H^,H^). Yes, if H doesn't abort its simulation, then for THAT H^, we have that H^(H^) is non-halting, but so is H(H^,H^), and thus THAT H / H^ pair fails to be a counter example

When you program H to abort its simulation of H^ at some point, and build your H^ on that H, then H(H^,H^), will return the non-halting answer, and H^(H^) when PROPERLY run or simulated halts, because H has the same "cut off" logic at the factorial above.

The naive expansion thinks it is infinite, but the correct expansion sees the cut off and sees that it is actually finite.
----------------------------------------------------------------------

A good symbolic manipulation system or a theorem prover with appropriate axioms and rules of inference could surely handle forms such as Fact(n+1)/fact(n) without breathing hard. It is only you, an ignorant fool, who seems to think that the unthinking infinite unrolling of a form must occur. Only you would think that a solver system would completely unroll a form before analyzing it and applying transformations to it.

Son, it don't work that way (unless you are defining and making a mess trying to write the system yourself). Systems usually have rules that make small incremental transformations and usually search breadth first with perhaps a limited amount of depth first interludes. If they don't use a breadth first strategy, they will not be able to claim the completeness property. (See resolution theorem prover literature for some explanation. You wont understand it but you can cite as if you did!)

Richard was trying to explain this to you in the snipped portion I recited just above. Question for Peter holding his pecker: How do you always and I mean always manage to delete the part of a message you respond too that addresses the point you now try to make?

A typically subsequence you might see in the trace: would include in order but not necessarily consecutively:
    Fact(n+1)/Fact(n)
    (n+1)*Fact(n)/Fact(n)
    (n+1)
Some interspersed terms such as Fact(n+1)/(n*Fact(n-1)) would be found too. In some circumstances, these other terms might be helpful. A theorem prover or manipulator does all of this, breadth first, hoping to blindly stumble on a solution. You can provide heuristics that might speed up the process but no advice short of an oracle will get you even one more result. (Another manifestation of HP.) It's the slow grinding through the possibilities that guarantees that if a result can be found, it will be found. And all the theory that you don't understand says that's the best you can do.

Ben and I disagree on reasons for your type of total dishonesty. He thinks that you are so self deluded that you actual believe what you are saying; that you are so self deluded that the dishonest utterances are just your subconscious protecting your already damaged ego. To that, I say phooey; you are just a long term troll who lies a lot about math, about your history and health, and about your accomplishments.

I don't believe that you will read this before you start to respond but that's okay. Understanding is not required. Neither is respect in either direction.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer

On 5/1/2022 6:12 AM, Richard Damon wrote:

On 5/1/22 6:58 AM, olcott wrote:

On 5/1/2022 4:24 AM, Mikko wrote:

On 2022-04-30 18:15:19 +0000, Aleksy Grabowski said:

I just want to add some small note.

This "not(true(_))" thing is misleading.
Please note that it is *not* a negation.
Functionally it is equivalent to:

X = foo(bar(X)).

That's correct. Prolog language does not give any inherent semantics to
data structures. It only defines the execution semantics of language
structures and standard library symbols. Those same synbols can be used
in data structures with entirely different purposes.

Mikko

negation, not, \+
The concept of logical negation in Prolog is problematical, in the sense that the only method that Prolog can use to tell if a proposition is false is to try to prove it (from the facts and rules that it has been told about), and then if this attempt fails, it concludes that the proposition is false. This is referred to as negation as failure.

http://www.cse.unsw.edu.au/~billw/dictionaries/prolog/negation.html
This is actually a superior model to convention logic in that it only seeks to prove not true, thus detects expressions of language that are simply not truth bearers.


Expressions of (formal or natural) language that can possibly be resolved to a truth value are [truth bearers].

There are only two ways that an expression of language can be resolved to a truth value:
(1) An expression of language is assigned a truth value such as "cats are animals" is defined to be true.

(2) Truth preserving operations are applied to expressions of language that are known to be true. {cats are animals} and {animals are living things} therefore {cats are living things}. Copyright 2021 PL Olcott

So you are sort of answering your own question. The model of logic that Prolog handles isn't quite the same as "conventional" logic, in part due to the way it (doesn't) define Logical Negation.

This seems to fit into your standard misunderstanding of things.

Prolog has a better model in that it can detect semantic paradoxes.
LP ↔ ¬True(LP) is correctly assessed as neither true nor false.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer

On 5/1/2022 6:18 AM, Richard Damon wrote:

On 4/30/22 11:15 PM, olcott wrote:

On 4/30/2022 10:11 PM, Richard Damon wrote:

On 4/30/22 10:56 PM, olcott wrote:

On 4/30/2022 9:38 PM, Richard Damon wrote:

On 4/30/22 10:21 PM, olcott wrote:

On 4/30/2022 9:00 PM, Richard Damon wrote:

On 4/30/22 9:42 PM, olcott wrote:

On 4/30/2022 8:08 PM, Richard Damon wrote:

On 4/30/22 3:02 AM, olcott wrote:

LP := ~True(LP) is translated to Prolog:

?- 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)

https://www.researchgate.net/publication/350789898_Prolog_detects_and_rejects_pathological_self_reference_in_the_Godel_sentence

Since it isn't giving you a "syntax error", it is probably correct Prolog. Not sure if your interpretation of the results is correct.

All that false means is that the statement


LP = not(true(LP))

is recursive and that Prolog can't actually evaluate it due to its limited logic rules.

That is not what Clocksin & Mellish says. They say it is an erroneous "infinite term" meaning that it specifies infinitely nested definition like this:

No, that IS what they say, that this sort of recursion fails the test of Unification, not that it is has no possible logical meaning.

Prolog represents a somewhat basic form of logic, useful for many cases, but not encompassing all possible reasoning systems.

Maybe it can handle every one that YOU can understand, but it can't handle many higher order logical structures.

Note, for instance, at least some ways of writing factorial for an unknown value can lead to an infinite expansion, but the factorial is well defined for all positive integers. The fact that a "prolog like" expansion operator might not be able to handle the definition, doesn't mean it doesn't have meaning.

It is really dumb that you continue to take wild guesses again the verified facts.

Please read the Clocksin & Mellish (on page 3 of my paper) text and eliminate your ignorance.

I did. You just don't seem to understand what I am saying because it is above your head.

Prolog is NOT the defining authority for what is a valid logical statement, but a system of programming to handle a subset of those statements (a useful subset, but a subset).

The fact that Prolog doesn't allow something doesn't mean it doesn't have a logical meaning, only that it doesn't have a logical meaning in Prolog.

In this case it does. I have spent thousands of hours on the semantic error of infinitely recursive definition and written a dozen papers on it. Glancing at one of two of the words of Clocksin & Mellish does not count as reading it.

And it appears that you don't understand it, because you still make category errors when trying to talk about it.

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 so on. So Y ends up standing for some kind of infinite structure.
END:(Clocksin & Mellish 2003:254)

foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(...)))))))))))))))

Right. but some infinite structures might actually have meaning.

Not in this case, it is very obvious that no theorem prover can possibly prove any infinite expression. It is the same thing as a program that is stuck in an infinite loop.

As Jeff pointed out, your claim is shown false, that some statements with infinite expansions can be worked with by some automatic solvers.

That is the dishonest dodge of the strawman error. The particular expression at hand is inherently incorrect and thus any system that proves it is a broken system.

This just proves that you don't really understand the effects of "recursion" and "self-reference", and perhaps a source of your error in trying to reason about thing like this Halting Problem proof or Godels Incompleteness Theory and his expression "G".

foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(...))))))))))))
The "..." indicates infinite expansion, thus this expression can never be proved by any system.

If something that is obviously undoable is your mind is shown to be in fact doable, the problem isn't in the ability to do it, but in the ability of your mind to understand. Your misconceptions don't change the nature of reality, but reality proves you wrong.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer

On 5/1/2022 6:26 AM, Richard Damon wrote:

On 5/1/22 7:06 AM, olcott wrote:

On 5/1/2022 4:38 AM, Mikko wrote:

On 2022-04-30 20:48:47 +0000, olcott said:

On 4/30/2022 4:31 AM, Mikko wrote:

On 2022-04-30 07:02:23 +0000, olcott said:

LP := ~True(LP) is translated to Prolog:

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

This is correct but to fail would also be correct.

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

unify_with_occurs_check must fail if the unified data structure
would contain loops.

Mikko

The above is the actual execution of actual Prolog code using
(SWI-Prolog (threaded, 64 bits, version 7.6.4).

Another Prolog implementation might interprete LP = not(true(LP)) differently
and still conform to the prolog standard.

According to Clocksin & Mellish it is not a mere loop, it is an "infinite term" thus infinitely recursive definition.

When discussing data structures, "infinite" and "loop" mean the same.
The data structure is infinitely deep but contains only finitely many
distinct objects and occupies only a finite amount of memory.

That is incorrect. any structure that is infinitely deep would take all of the memory that is available yet specifies an infinite amount of memory.

Nope, a tree that one branch points into itself higher up represents a tree with infinite depth, but only needs a finite amount of memory. Building such a structure may require the ability to forward declare something or reference something not yet defined.

That is counter-factual. unify_with_occurs_check determines that it would require infinite memory and then aborts its evaluation.

foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(...))))))))))))
"..." indicates infinite depth, thus infinite string length.

Some infinities have finite representation. You don't seem able to understand that.

Yes, some naive ways of expanding them fail, but the answer to that is you just don't do that, but need to use a less naive method.

I am trying to validate whether or not my Prolog code encodes the Liar Paradox.

That cannot be inferred from Prolog rules. Prolog defines some encodings
like how to encode numbers with characters of Prolog character set but for
more complex things you must make your own encoding rules.

Mikko

This says that G is logically equivalent to its own unprovability in F
G ↔ ¬(F ⊢ G) and fails unify_with_occurs_check when encoded in Prolog.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer

On 5/1/22 7:28 AM, olcott wrote:

On 5/1/2022 4:45 AM, Mikko wrote:

On 2022-05-01 03:15:56 +0000, olcott said:

Not in this case, it is very obvious that no theorem prover can possibly prove any infinite expression.

Doesn't matter as you can't give an infinite expression to a theorem
prover.

Mikko

*You can and Prolog can detect and reject it*

Which just shows PROLOG can't handle that sort of expression, not that it logically doens't have a meaning.

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 so on. So Y ends up standing for some kind of infinite structure.

<inserted for clarity>
foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(...))))))))))))
</inserted for clarity>

Note that, whereas they may allow you to construct something like this, most Prolog systems will not be able to write it out at the end. According to the formal definition of Unification, this kind of “infinite term” should never come to exist. Thus Prolog systems that allow a term to match an uninstantiated subterm of itself do not act correctly as Resolution theorem provers. In order to make them do so, we would have to add a check that a variable cannot be instantiated to something containing itself. Such a check, an occurs check, would be straightforward to implement, but would slow down the execution of Prolog programs considerably. Since it would only affect very few programs, most implementors have simply left it out 1.

1 The Prolog standard states that the result is undefined if a Prolog system attempts to match a term against an uninstantiated subterm of itself, which means that programs which cause this  to happen will not be portable. A portable program should ensure that wherever an occurs check might be applicable the built-in predicate unify_with_occurs_check/2 is used explicitly instead of the normal unification operation of the Prolog implementation. As its name suggests, this predicate acts like =/2 except that it fails if an occurs check detects an illegal attempt to instantiate a variable.
END:(Clocksin & Mellish 2003:254)

Clocksin, W.F. and Mellish, C.S. 2003. Programming in Prolog Using the ISO Standard Fifth Edition, 254. Berlin Heidelberg: Springer-Verlag.

So all Clocksin & Melish is saying is that such an expression fails in PROLOG, not that it doesn't have a valid logical meaning.

The world is NOT Prolog, and I suspect Prolog isn't sufficient to handle the logic needed to process the Godel Sentence, so can't be used to disprove it.

On 5/1/22 7:45 AM, olcott wrote:

On 5/1/2022 6:12 AM, Richard Damon wrote:

On 5/1/22 6:58 AM, olcott wrote:

On 5/1/2022 4:24 AM, Mikko wrote:

On 2022-04-30 18:15:19 +0000, Aleksy Grabowski said:

I just want to add some small note.

This "not(true(_))" thing is misleading.
Please note that it is *not* a negation.
Functionally it is equivalent to:

X = foo(bar(X)).

That's correct. Prolog language does not give any inherent semantics to
data structures. It only defines the execution semantics of language
structures and standard library symbols. Those same synbols can be used
in data structures with entirely different purposes.

Mikko

negation, not, \+
The concept of logical negation in Prolog is problematical, in the sense that the only method that Prolog can use to tell if a proposition is false is to try to prove it (from the facts and rules that it has been told about), and then if this attempt fails, it concludes that the proposition is false. This is referred to as negation as failure.

http://www.cse.unsw.edu.au/~billw/dictionaries/prolog/negation.html
This is actually a superior model to convention logic in that it only seeks to prove not true, thus detects expressions of language that are simply not truth bearers.


Expressions of (formal or natural) language that can possibly be resolved to a truth value are [truth bearers].

There are only two ways that an expression of language can be resolved to a truth value:
(1) An expression of language is assigned a truth value such as "cats are animals" is defined to be true.

(2) Truth preserving operations are applied to expressions of language that are known to be true. {cats are animals} and {animals are living things} therefore {cats are living things}. Copyright 2021 PL Olcott

So you are sort of answering your own question. The model of logic that Prolog handles isn't quite the same as "conventional" logic, in part due to the way it (doesn't) define Logical Negation.

This seems to fit into your standard misunderstanding of things.

Prolog has a better model in that it can detect semantic paradoxes.
LP ↔ ¬True(LP) is correctly assessed as neither true nor false.

"Better" is as subjective word unless you define an objective criteria. The fact that Prolog doesn't have the expresiability to actually write the Godel sentence, means it can't actually be used to disprove it.

Misusing notations to show something is invalid logic.

I don't know Prolog well enough, but if the statement doesn't actaully mean what you want it to mean to Prolog (as others have said), then the fact that Prolog gives you the answer you want doesn't mean anything.