Hi
I'm running virtuoso 6.1.3127 on Mac Lion (Open Source version).
I can get all types of inferencing I need to work, except fro rdfs:range
and rdfs:domain type inference.
Here's the relevant snippet from my RDF/XML
<rdf:RDF xmlns="http://www.w3.org/1999/02/22-rdf-syntax-ns#";
xmlns:owl="http://www.w3.org/2002/07/owl#";
xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#";
xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#";
xmlns:ex="http://www.example.com/example#";
xmlns:err="http://www.w3.org/2005/xqt-errors#";
xmlns:fn="http://www.w3.org/2005/xpath-functions#";
xmlns:xs="http://www.w3.org/2001/XMLSchema#";>
<rdf:Description rdf:about="http://www.example.com/example#ian";>
<ex:has-pet rdf:resource="http://www.example.com/example#slate"/>
</rdf:Description>
<rdf:Description rdf:about="http://www.example.com/example#fhas-pet";>
<rdf:type rdf:resource="http://www.w3.org/2002/07/owl#ObjectProperty"/>
</rdf:Description>
<rdf:Description rdf:about="http://www.example.com/example#has-pet";>
<rdfs:range rdf:resource="http://www.example.com/example#pet"/>
</rdf:Description>
<rdf:Description rdf:about="http://www.example.com/example#has-pet";>
<rdfs:domain rdf:resource="http://www.example.com/example#human"/>
</rdf:Description>
</rdf:RDF>
I should be able to infer that ian is of type human and slate is of type
pet .... but it just returns nothing
I'm using the sparql querty
select * where {ex:ian rdf:type ?y}
what am I missing?
N.B. subClassOf,subPropertyOf, inverseOf, symmetric and transitive
property inferecing works on the same dataset ... just not the
range/domain inferemcing queries
ian
--
Ian Harrison Email: [email protected]
