org.swrlapi.parser.SWRLParseException: Invalid SWRL atom predicate 'Person'

[ChristophB]

Hello,

i get an org.swrlapi.parser.SWRLParseException when i try to create an SWRLRule.

Minimal example:

private static final String ONT_IRI = "http://example.org/example#";

// ...

OWLOntologyManager manager = OWLManager.createOWLOntologyManager();
OWLDataFactory factory = manager.getOWLDataFactory();
OWLOntology ontology = manager.createOntology(IRI.create(ONT_IRI));

manager.addAxiom(ontology, factory.getOWLDeclarationAxiom(factory.getOWLDataProperty(IRI.create(ONT_IRI, "hasAge"))));
manager.addAxiom(ontology, factory.getOWLDeclarationAxiom(factory.getOWLClass(IRI.create(ONT_IRI, "Person"))));
manager.addAxiom(ontology, factory.getOWLDeclarationAxiom(factory.getOWLClass(IRI.create(ONT_IRI, "Adult"))));

SWRLRuleEngine ruleEngine = SWRLAPIFactory.createSWRLRuleEngine(ontology);
SWRLAPIRule rule = ruleEngine.createSWRLRule(
    "IsAdult-Rule",
    "Person(?p) ^ hasAge(?p, ?age) ^ swrlb:greaterThan(?age, 17) -> Adult(?p)"
); // Exception is thrown here

It seams that the SWRLRuleEngine does not recognize the Person class.

Dependencies:

• edu.stanford.swrl.swrlapi v2.0.4
• edu.stanford.swrl.swrlapi-drools-engine v2.0.4
• net.sourceforge.owlapi.owlapi-api v4.2.7

Do you have any hints to resolve this issue?

Thanks, Christoph

[martinjoconnor]

If you will need to set a default namespace for the ontology (in this case http://example.org/example#) in order to use the unqualified name.

[ChristophB]

Thanks, manager.getOntologyFormat(ontology).asPrefixOWLOntologyFormat().setDefaultPrefix(ONT_IRI); solved the problem.