- 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