Re: "is a" & "has a" relations re: 2nd-Order Logic


Karmata wrote:
FYI: check out this pattern I just noticed. The relation of a
first-order object (FOO) to a first-order predicate (FOP) seems to be
an "is a" relation. But the relation of a FOO to a second-order
predicate (SOP) seems to be a "has a" relation. At least in these
examples:

SOP: shape

Shape IS NOT a 2nd-order predicate.
I don't know WHO told you this but whoEVER it is,
you have been SERIOUSLY misled.

FOP: cube
FOO: x
"x is a cube." "x has a shape."

x is NOT a "first-order object" in this context.
The objects/arguments of first-order things are ZEROth
order things, just as the arguments of 2nd-order predicates
are 1st-order predicates.

The fact that x has a shape makes shape a FUNCTION,
NOT a predicate. x has to have some PARTICULAR shape,
in order to have a shape (in this case, it's cube-shaped).
In formal language, you would say something like
shape(x)=cube. This is NOT the same cube as the one
you get from the predicate Cube(x). Cube(x) IS a predicate
because the ONLY values it can return are true and false.
Shape(x) is NOT a predicate.

If you are considering all the possible properties that
0th-order objects might have and are noticing that some of
them are about its shape, while others are about other attributes,
then, yes, you can get *a* 2nd-order predicate that way.
But it is going to get you hopelessly confused with other uses.

SOP: flavor
FOP: sweet
FOO: x
"x is a sweet thing."

For that, you just say "x is sweet".

"x has a flavor."

That, again, implies that flavor(x) IS A FUNCTION, NOT
a predicate. But of course, you could apply flavor to
sweet instead of to x and get a whole different intension.
It DOES MATTER what type of thing the ARGUMENT is.
flavor(x) and flavor(sweet) are two completely DIFFERENT
KINDS of things. Under normal circumstances they are not
BOTH even GRAMMATICAL.


SOP: mood
FOP: happy
FOO: x
"x is a happy boy." "x has a mood."

You are missing the one that COUNTS, namely,
"happy is a mood", "sweet is a flavor", "cube
(or cubical) is a shape".
What anything "has" is just irrelevant, since
you can always define a class of things such that
being one of those things equates to having certain
attributes.

.

Relevant Pages

  • Re: "is a" & "has a" relations re: 2nd-Order Logic
    ... But the relation of a FOO to a second-order ... predicate (SOP) seems to be a "has a" relation. ... FOP: cube ...
    (sci.logic)
  • Re: backtracking in prolog
    ... I want to create a predicate, which takes two arguments, each which is ... The predicate is called foo for example (so foo takes two ... You seem to want the backtracking to continue only ... that does not satisfy the given predicate. ...
    (comp.lang.prolog)
  • Re: swi-prolog non-dynamic predicates in thread local
    ... dynamic predicates (sometimes even different implementations of the ... In this case I'd have separate module for each thread and import each ... But when I'd like to have module 'bb' with shared predicate imported ... from 'a:tst' redefined foo predicate from file c.pl: ...
    (comp.lang.prolog)