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devel / comp.lang.prolog / Is this correct Prolog?

SubjectAuthor
* Is this correct Prolog?olcott
+* Is this correct Prolog?Mikko
|+* Is this correct Prolog?Aleksy Grabowski
||+* Is this correct Prolog?olcott
|||`* Is this correct Prolog?Mikko
||| `- Is this correct Prolog?olcott
||`* Is this correct Prolog?Mikko
|| `* Is this correct Prolog?olcott
||  `* Is this correct Prolog?Richard Damon
||   `* Is this correct Prolog?olcott
||    `* Is this correct Prolog?Richard Damon
||     `* Is this correct Prolog?olcott
||      `- Is this correct Prolog?Richard Damon
|`* Is this correct Prolog?olcott
| `* Is this correct Prolog?Mikko
|  `* Is this correct Prolog?olcott
|   `* Is this correct Prolog?Richard Damon
|    `* Is this correct Prolog?olcott
|     `* Is this correct Prolog?Richard Damon
|      `* Is this correct Prolog?olcott
|       `- Is this correct Prolog?Richard Damon
`* Is this correct Prolog?Richard Damon
 +* Is this correct Prolog?olcott
 |`* Is this correct Prolog?Richard Damon
 | `* Is this correct Prolog?olcott
 |  `* Is this correct Prolog?Richard Damon
 |   `* Is this correct Prolog?olcott
 |    `* Is this correct Prolog?Richard Damon
 |     `* Is this correct Prolog?olcott
 |      +* Is this correct Prolog?Jeff Barnett
 |      |+* Is this correct Prolog?olcott
 |      ||`- Is this correct Prolog?Richard Damon
 |      |`* Is this correct Prolog?Mr Flibble
 |      | +- Is this correct Prolog?polcott
 |      | +- Is this correct Prolog?Richard Damon
 |      | `- Is this correct Prolog?Jeff Barnett
 |      +* Is this correct Prolog?Mikko
 |      |`* Is this correct Prolog?olcott
 |      | `* Is this correct Prolog?Richard Damon
 |      |  `* Is this correct Prolog?olcott
 |      |   `- Is this correct Prolog?Richard Damon
 |      `* Is this correct Prolog?Richard Damon
 |       `* Is this correct Prolog?olcott
 |        `- Is this correct Prolog?Richard Damon
 `* Is this correct Prolog?olcott
  `* Is this correct Prolog?olcott
   +* Is this correct Prolog?olcott
   |`* Is this correct Prolog?olcott
   | `- Is this correct Prolog?olcott
   `* Is this correct Prolog?olcott
    `- Is this correct Prolog?André G. Isaak

Pages:123
Is this correct Prolog?

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 by: olcott - Sat, 30 Apr 2022 07:02 UTC

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

Re: Is this correct Prolog?

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Subject: Re: Is this correct Prolog?
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 by: Mikko - Sat, 30 Apr 2022 09:31 UTC

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

Re: Is this correct Prolog?

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From: hur...@gmail.com (Aleksy Grabowski)
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Subject: Re: Is this correct Prolog?
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 by: Aleksy Grabowski - Sat, 30 Apr 2022 18:15 UTC

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)).

On 4/30/22 11:31, 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
>

--
Alex

Re: Is this correct Prolog?

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 by: olcott - Sat, 30 Apr 2022 20:48 UTC

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).

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

I am trying to validate whether or not my Prolog code encodes the Liar
Paradox. I believe that it does and it also shows exactly how the Liar
Paradox is erroneous.

From all of my analysis and research this expression correctly
formalizes the Liar Paradox: LP := ~True(LP)

--
Copyright 2022 Pete Olcott

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

Re: Is this correct Prolog?

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 by: olcott - Sat, 30 Apr 2022 21:08 UTC

On 4/30/2022 1:15 PM, Aleksy Grabowski wrote:
> 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)).
>

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

> On 4/30/22 11:31, 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.

Not according to Clocksin & Mellish, unify_with_occurs_check detects an
"infinite term" which is the same thing as infinititely recursive
definition, thus not at all any sort of "do while loop".

LP := ~True(LP)
specifies ~True(~True(~True(~True(~True(..)))))

>>
>> Mikko
>>
>

--
Copyright 2022 Pete Olcott

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

Re: Is this correct Prolog?

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 by: Richard Damon - Sun, 1 May 2022 01:08 UTC

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.

Re: Is this correct Prolog?

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 by: olcott - Sun, 1 May 2022 01:42 UTC

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

Re: Is this correct Prolog?

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 by: olcott - Sun, 1 May 2022 01:47 UTC

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

Re: Is this correct Prolog?

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 by: Richard Damon - Sun, 1 May 2022 02:00 UTC

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.
>
>

Re: Is this correct Prolog?

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 by: olcott - Sun, 1 May 2022 02:21 UTC

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

Re: Is this correct Prolog?

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 by: Richard Damon - Sun, 1 May 2022 02:38 UTC

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.

Re: Is this correct Prolog?

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 by: olcott - Sun, 1 May 2022 02:56 UTC

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

Re: Is this correct Prolog?

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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Sun, 1 May 2022 03:11 UTC

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.

Re: Is this correct Prolog?

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 by: olcott - Sun, 1 May 2022 03:15 UTC

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

Re: Is this correct Prolog?

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 by: olcott - Sun, 1 May 2022 04:49 UTC

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

Re: Is this correct Prolog?

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 by: Jeff Barnett - Sun, 1 May 2022 05:24 UTC

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.

Click here to read the complete article

Re: Is this correct Prolog?

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 by: Mikko - Sun, 1 May 2022 09:24 UTC

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

Re: Is this correct Prolog?

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 by: Mikko - Sun, 1 May 2022 09:26 UTC

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

Re: Is this correct Prolog?

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 by: Mikko - Sun, 1 May 2022 09:38 UTC

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.

> 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

Re: Is this correct Prolog?

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 by: Mikko - Sun, 1 May 2022 09:45 UTC

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

Re: Is this correct Prolog?

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 by: olcott - Sun, 1 May 2022 10:58 UTC

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

Re: Is this correct Prolog?

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 by: olcott - Sun, 1 May 2022 11:00 UTC

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

Re: Is this correct Prolog?

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 by: olcott - Sun, 1 May 2022 11:06 UTC

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

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Subject: Re: Is this correct Prolog?
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 by: Richard Damon - Sun, 1 May 2022 11:12 UTC

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.

Re: Is this correct Prolog?

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Subject: Re: Is this correct Prolog?
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 by: Richard Damon - Sun, 1 May 2022 11:18 UTC

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.


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