Re: RDFS interpretation of typed literals

>Considering the RDF statement:
>
>   ex:s ex:p "foo"^^ddd .
>
>where ddd is any URI other than rdf:XMLLiteral.
>
>Is this statement true, false or undefined in an RDF[S] interpretation?
>
>As far as I can tell, the denotation of "foo"^^ddd is defined by 
>interpretation function IL, for which there are no semantic 
>constraints in RDF and RDFS interpretations except for the datatype 
>rdf:XMLLiteral.
>
>So I think the above statement is either True or False, depending on 
>the interpretation used.  So far, so good.  But is this an 
>RDFS-entailment:
>
>   ex:s ex:p "foo"^^ddd .
>|=
>   ex:s ex:p _:b .
>   _:b rdf:type rdfs:Literal .
>
>Intuitively, I would have said "yes", but as far as I can tell from 
>the semantics spec, the answer is "no".  This is an entailment only 
>in a {RDFS,ddd}-interpretation, because it is only a 
>D-interpretation that places the required constraints on the 
>relationships between I(rdfs:Literal), LV and IL.
>
>Am I correct?

Yes. Using a typed literal with an unknown type (which means any 
except rdf:XMLLiteral in bare RDFS) is hostage to fortune. It denotes 
*something*, but you only know its a literal value when you know that 
the string is well-formed for that dtype: and if the dtype is 
unknown, then you don't know that (yet).

Pat

>Is this what we expected?
>
>#g
>--
>
>PS:  this would be consistent with the entailment rules, for which I 
>was working through my implementation when I noticed this.
>
>
>------------
>Graham Klyne
>For email:
>http://www.ninebynine.org/#Contact


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Received on Tuesday, 16 December 2003 09:32:45 UTC