RDFS: How to subclass a FOAF class

599 views Asked by At

Is it possible to subclass a class found in FOAF (http://xmlns.com/foaf/spec/)? I tried something like the code below, but I am not sure if it is the proper way to do it or not.

<rdfs:Class rdf:ID="user"> 
    <rdfs:subClassOf rdf:resource="http://xmlns.com/foaf/0.1/#Agent" />
    <rdfs:comment> 
        The class of users, subclass of foaf:Agent.
    </rdfs:comment>  
</rdfs:Class>
1

There are 1 answers

2
Joshua Taylor On BEST ANSWER

Your snippet, although not a complete RDF document, is the right way to make yourdoc#user a subclass of http://xmlns.com/foaf/0.1/#Agent. However, that latter class is not the FOAF Agent class. The FOAF Agent class is identified by the URI http://xmlns.com/foaf/0.1/Agent (with no #). It might be useful to have a look at the actual FOAF ontology, because you can see how it defines subclasses of Agent. For instance, it declared foaf:Organization with

<rdfs:Class rdf:about="http://xmlns.com/foaf/0.1/Organization" rdfs:label="Organization" rdfs:comment="An organization." vs:term_status="stable">
  <rdf:type rdf:resource="http://www.w3.org/2002/07/owl#Class"/>
  <rdfs:subClassOf rdf:resource="http://xmlns.com/foaf/0.1/Agent"/>
  <rdfs:isDefinedBy rdf:resource="http://xmlns.com/foaf/0.1/"/>
  <owl:disjointWith rdf:resource="http://xmlns.com/foaf/0.1/Person"/>
  <owl:disjointWith rdf:resource="http://xmlns.com/foaf/0.1/Document"/>
</rdfs:Class>

If you're writing this by hand, it's much easier to work in the Turtle or N3 serialization, where that would be:

foaf:Organization  a      rdfs:Class , owl:Class ;
        rdfs:comment      "An organization." ;
        rdfs:isDefinedBy  foaf: ;
        rdfs:label        "Organization" ;
        rdfs:subClassOf   foaf:Agent ;
        owl:disjointWith  foaf:Person , foaf:Document ;
        vs:term_status    "stable" .