- From: Patrick Stickler <patrick.stickler@nokia.com>
- Date: Tue, 30 Apr 2002 12:28:14 +0300
- To: <www-archive@w3.org>
------ Forwarded Message
From: Patrick Stickler <patrick.stickler@nokia.com>
Date: Tue, 30 Apr 2002 09:58:50 +0300
To: ext Dan Brickley <danbri@w3.org>
Cc: Brian McBride <brian_mcbride@hp.com>, Pat Hayes <phayes@ai.uwf.edu>, ext
Sergey Melnik <melnik@DB.Stanford.EDU>
Subject: Specific questions regarding rdfs:Literal, possible closure...
Dan (et. al.),
I would appreciate some answers and clarifications to the following
specific questions regarding literals and rdfs:Literal:
My own understanding (how I would answer these questions) are
provided in square brackets.
1. Are all strings denoted by literal strings in the graph automatically
members of the class rdfs:Literal?
[Yes]
2. Are all members of rdfs:Literal also members of rdf:Resource? I.e.
is rdfs:Literal rdfs:subClassOf rdf:Resource?
[No]
3. Can a blank node or uriref denote a literal string and appear in
the graph wherever that literal string might appear?
[Yes]
4. Is the object of the following statement a member of rdfs:Literal?
x:foo rdfs:comment "This is a foo." .
[Yes]
Is the object a literal node?
[Yes]
5. Is the object of the following statement a member of rdfs:Literal?
x:foo rdfs:comment _:y .
[Yes, per rdfs:comment rdfs:range rdfs:Literal]
Is the object a literal node?
[No]
--
If the above answers are correct, then:
1. The addition of the following closure rule, IMO, resolves my objection
to the present datatyping specification:
IF
?p rdfd:datatype ?d .
?s ?p ?l .
?l rdf:type rdfs:Literal .
THEN
?s ?p ?x .
?x rdfd:lex ?l .
and:
2. The following 'template' constrains the property values to the
inline idiom and the lexical forms of a given datatype:
?p rdfd:datatype ?d .
?p rdfs:range rdfs:Literal .
3. The following 'template' constrains the property values to the
lexical form and datatype property idioms and the datatype values
of a given datatype:
?p rdfd:datatype ?d .
?p rdfs:range ?d .
4. Conflicts with daml:uniqueProperty will remain when both the
inline idiom and either the lexical form or datatype property
idioms are used for such a property in the same graph. This
is simply due to the fact that a lexical form will never equal
a datatype value, and one may specify either as the object of
a property -- unless one uses one of the constraints as defined
in #2 or #3 above.
Patrick
--
Patrick Stickler Phone: +358 50 483 9453
Senior Research Scientist Fax: +358 7180 35409
Nokia Research Center Email: patrick.stickler@nokia.com
------ End of Forwarded Message
Received on Tuesday, 30 April 2002 05:25:09 UTC