XPost: comp.theory, comp.lang.prolog

On 5/1/2022 9:39 AM, Ben wrote:

Mr Flibble <[email protected]> writes:

Recursive definitions are fine, infinitely recursive definitions (such
as The Halting Problem) are INVALID.

(1) The halting problem is not an infinitely recursive definition.

(2) Infinitely recursive definitions are often fine. For example, the
list of Fibonacci numbers:

This type is never fine: // Adapted from Clocksin & Mellish foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(...))))))))))))

Because Gödel says
14 Every epistemological antinomy can likewise be used for a similar undecidability proof

G ↔ ¬Provable(F, G) is an epistemological antinomy therefore it is necessarily sufficiently equivalent to his G.

Likewise with this one: LP ↔ ¬True(LP)
It can be evaluated as semantically incorrect without the need to define True(). No matter how True() is defined LP ↔ ¬True(LP) is semantically incorrect because it specifies this:

LP ↔ ¬True(¬True(¬True(¬True(¬True(¬True(¬True(¬True(LP))))))))
and Prolog can detect and reject this with unify_with_occurs_check.

fibs = 1 : 1 : zipWith (+) fibs (tail fibs)

or the list of factorials:

facts = prod 1 1 where prod p n = p : prod (p*n) (n+1)

--
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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XPost: comp.theory, comp.lang.prolog

On 5/1/22 12:05 PM, polcott wrote:

On 5/1/2022 9:39 AM, Ben wrote:

Mr Flibble <[email protected]> writes:

Recursive definitions are fine, infinitely recursive definitions (such
as The Halting Problem) are INVALID.

(1) The halting problem is not an infinitely recursive definition.

(2) Infinitely recursive definitions are often fine.  For example, the
list of Fibonacci numbers:

This type is never fine: // Adapted from Clocksin & Mellish foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(...))))))))))))

Depends on the definition of foo.

If foo is the function "fact" I have presented, then the unwinding
becomes exactly that to a too dumb naive expansion.

Thus "never" is incorrect.

Because Gödel says
14 Every epistemological antinomy can likewise be used for a similar undecidability proof

G ↔ ¬Provable(F, G) is an epistemological antinomy therefore it is necessarily sufficiently equivalent to his G.

Likewise with this one: LP ↔ ¬True(LP)
It can be evaluated as semantically incorrect without the need to define True(). No matter how True() is defined LP ↔ ¬True(LP) is semantically incorrect because it specifies this:

LP ↔ ¬True(¬True(¬True(¬True(¬True(¬True(¬True(¬True(LP))))))))
and Prolog can detect and reject this with unify_with_occurs_check.

Because Prolog is too limited to be able to fully handle this sort of recursion. Prolog apparently can't handle the recursive definition of
fact(n), which just proves that its rejection of something as
"infinitely recursive" is NOT "Proof" that it is invalid.

   fibs = 1 : 1 : zipWith (+) fibs (tail fibs)

or the list of factorials:

    facts = prod 1 1 where prod p n = p : prod (p*n) (n+1)

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