SWRL rule on n-ary relation in OWL

Question

I want to use OWL (in Protege) to encode the ternary relation ancestorOf(x, y, p); saying that y is the ancestor of x with probability p. Because object properties only support binary relations, my relation has to be represented as a Relation class and a relation individual, with relations to x, y and p (like in this design pattern).

I really do not know how to write an SWRL rule to infer a transitive relation. I.e. that

ancestorOf(x, y, p1) ^ ancestorOf(y, z, p2) -> ancestorOf(x, z, p1 * p2)

I would be thankful is somebody could point me in the correct direction.

Thanks in advance!

score 2 · Accepted Answer · answered Mar 21 '18 at 01:55

The first thing to do is to ensure your ontology is correctly designed for the task. As you rightly pointed out you will need reification to achieve this. This means you will need to introduce a class for representing your n-ary relation, i.e.:

The related ontology with rules and some individuals to test with will look as follows:

ObjectProperty: ancestor
    Domain: AncestorProbability
    Range: Person

ObjectProperty: descendant
    Domain: AncestorProbability
    Range: Person

DataProperty: probability
    Domain: AncestorProbability
    Range: xsd:decimal

Class: AncestorProbability

Class: Person

Individual: _a1
    Types: 
        AncestorProbability
    Facts:  
     ancestor  _y,
     descendant  _x,
     probability  0.2


Individual: _a2
    Types: 
        AncestorProbability
    Facts:  
     ancestor  _z,
     descendant  _y,
     probability  0.31
Individual: _a3
    Types: 
        AncestorProbability
    Facts: 
        descendant  _x

Individual: _x
    Types: Person

Individual: _y
    Types: Person

Individual: _z
    Types: Person

DifferentIndividuals: 
    _a1,_a2,_a3

DifferentIndividuals: 
    _x,_y,_z

Rule: 
    ancestor(?a1, ?y), descendant(?a1, ?x), probability(?a1, ?p1), 
    ancestor(?a2, ?z), descendant(?a2, ?y), probability(?a2, ?p2),
    descendant(?a3, ?x), swrlm:eval(?p3, "p1 * p2", ?p1, ?p2) 
    -> ancestor(?a3, ?z), probability(?a3, ?p3)

One important thing to note is that since there is no ancestorOf(x, y, p) property and you have to use reification, you have to specify your rule component-wise, where ancestor, descendant and probability are properties representing components of your AncestorProbability n-ary relation.

Another important thing to note is that descendant(?a3, ?x) must be added to indicate on which AncestorProbability individual the rule must be applied for which descendant.

One possible problem is that the reasoner you use may not support swrlm:eval(?p3, "p1 * p2", ?p1, ?p2), which is the problem I found when testing this with Protege and the HermiT reasoner. It is possible that some commercial reasoner does support this.

@JCR I am glad I could help! For interest sake what editor and reasoner are you using? — Henriette Harmse, Mar 21 '18 at 07:58
Very good answer. What kind of syntax are you using in your example? It looks much clearer than other owl-syntax variants which I am aware of. — cknoll, Oct 15 '20 at 11:14
@cknoll Thanks. This is [Manchester syntax](https://www.w3.org/TR/owl2-manchester-syntax/). — Henriette Harmse, Oct 15 '20 at 11:33

SWRL rule on n-ary relation in OWL

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