CODEWITHSUNDEEP
rdfsemantic-webowlontologyprotege
MiP
MiP

Reputation: 6462

Multiple owl:inverseOf with different domains or ranges

owl:inverseOf is often used to define inverse relations between properties. An axiom of the form P1 owl:inverseOf P2 asserts that for every pair (x,y) in the property extension of P1, there is a pair (y,x) in the property extension of P2, and vice versa for example hasChild and hasParent.

How can I define such inverse relationships between object properties with multiple domains and ranges, such as:

hasOwner owl:inverseOf hasDog
hasOwner owl:inverseOf hasCat

Classes: Person, Dog, Cat
ObjectProperties: hasOwner, hasDog, hasCat
    hasOwner:
        Domains: Dog or Cat
        Ranges: Person
    hasDog:
        Domains: Person
        Ranges: Dog
    hasCat:
        Domains: Person
        Ranges: Cat

If I known Mammy hasDog Spike and Mammy hasCat Tom, how can I model the ontology so that the reasoner can infer Spike hasOwner Mammy and Tom hasOwner Mammy?

Upvotes: 1

Views: 1143

Answers (2)

Antoine Zimmermann
Antoine Zimmermann

Reputation: 5515

Forgetting about the incomprehensible restriction that anything that has an owner is either a cat or a dog, you can get closer to a reasonable model with (in Turtle syntax):

hasDog rdfs:subPropertyOf [ owl:inverseOf hasOwner ] .
hasCat rdfs:subPropertyOf [ owl:inverseOf hasOwner ] .

This has the non-advantage of not creating a new class name or a new property name.

PS: I do know things that have an owner and are neither cats nor dogs.

Upvotes: 4

Sentry
Sentry

Reputation: 4113

I think there is a flaw in your logic.

hasOwner owl:inverseOf hasDog
hasOwner owl:inverseOf hasCat

These two statements can only be true if hasCat and hasDog are identical. Think about it, if P1 is the inverse of P2 and the inverse of P3, then by transitivity P2 must be the same as P3.

You should rather have a predicate hasPet as the inverse of hasOwner, create a class Pet and make Dog and Cat subclasses of it.

You could then make a rule to infer that from P1 hasPet P2 and P2 isA Cat follows P1 hasCat P2.

Upvotes: 1

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