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Re: Range interpretation question.

From: Jeen Broekstra <jbroeks@cs.vu.nl>
Date: Thu, 26 Apr 2001 16:22:31 +0200 (CEST)
To: Ken Baclawski <kenb@ccs.neu.edu>
cc: <www-rdf-logic@w3.org>, Arjohn Kampman <akam@aidministrator.nl>, <conen@gmx.de>
Message-ID: <Pine.GSO.4.32.0104261605490.18904-100000@flits.cs.vu.nl>
On Thu, 26 Apr 2001, Ken Baclawski wrote:

[snip axiom 17 not applicable to the class Literal]

> That can't be true.  If so, then there is a problem with
> this axiom:
> Ax23.    (=> (PropertyValue Object ?st ?o)
>           (and (Type ?st Statement)
>                (or (Type ?o Resource) (Type ?o Literal))))

I think that the problem is indeed with axiom 23, not 17.

In the axiomatic semantics, "Type" is explicitly defined as
a shorthand for RDF statements whose predicate is the
property rdf:type:

    Ax1. (<=> (Type ?s ?o) (PropertyValue type ?s ?o))

where "PropertyValue" is a ternary relation that provides a
one-on-one mapping between RDF statements and KIF relational
sentences (see section 2.1).

However, according to the RDF M&S spec:

    The property named "type" is defined to provide
    primitive typing. The formal definition of type is:

    5.There is an element of Properties known as RDF:type.
    6.Members of Statements of the form {RDF:type, sub, obj}
      must satisfy the following: sub and obj are members of
      Resources. [RDFSchema] places additional restrictions
      on the use of type.

Notice that according to point 6, the type relation can only
relate Resources. Not Literals. Thus, the rdf:type relation
can never be used to relate actual literals with the class
"rdfs:Literal", and thus an RDF statement of the form

    (type "someLiteral" rdfs:Literal)

can not be made. This in turn means that the
clause (Type ?o Literal) in axiom 23 is meaningless.

> How does one specify that ?o is a literal except by using
> (Type ?o Literal)?

I think a new relation is needed in the axiomatic semantics,
specifically for indicating that something is a Literal.
Since this knowledge is atomic, a unary relation would
probably suffice:

    (Literal ?o)

I'm no expert on KIF axioms, but this seems to me to solve
the problem.

Best regards,

                               Vrije Universiteit, Faculty of Sciences
Jeen Broekstra              Division of Mathematics & Computer Science
jbroeks@cs.vu.nl                                    de Boelelaan 1081a
http://www.cs.vu.nl/~jbroeks        1081 HV Amsterdam, the Netherlands
Received on Thursday, 26 April 2001 10:22:48 UTC

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