"Ignoring transitivity and/or complex subproperty axioms"

Question

I'm trying to obtain to model a property that is transitive, asymmetric and irreflexive. I understand that OWL reasoning does not support for this complex relationship.

However, is there a way to "trick" the reasoner? I want to use this for validation purposes, so I'm also trying SWRL rules (1), but the rule does not work (which makes sense, is the same reasoner).

(1) isBaseFor(?y1, ?y2) ^ isBaseFor(?y2, ?y3) -> isBaseFor(?y2, ?y3)

I'm thinking of using SQWRL, however this is a validation similar to SHACL or SPARQL (meaning that I have to run the validaton query separatly, which I'm trying to avoid).

Any ideas?

Did you not mean isBaseFor(?y1, ?y2) ^ isBaseFor(?y2, ?y3) -> isBaseFor(?y1, ?y3) -- ?y1 instead of ?y2 in rule head? — Henriette Harmse, Mar 22 '23 at 13:01
Thank you for the quick response! It works, I had a mistake in the rule... — António Antunes, Mar 22 '23 at 14:15

score 2 · Accepted Answer · answered Mar 22 '23 at 13:23

The "trick" is to create isBaseFor as irreflexive and asymmetric (without transitive characteristic) and add a SWRL Rule for the transitive characteristic. Here is the Manchester syntax:

ObjectProperty: isBaseFor
    Characteristics: 
    Irreflexive,
    Asymmetric
       
Individual: y1
    Facts:  
     isBaseFor  y2
    
Individual: y2
    Facts:  
     isBaseFor  y3
   
Individual: y3

    
DifferentIndividuals: 
    y1,y2,y3

Rule: 
    isBaseFor(?y1, ?y2), isBaseFor(?y2, ?y3) - isBaseFor(?y1, ?y3)

This will correctly give inconsistencies for:

isBaseFor(y1, y2) and isBaseFor(y2, y1) because of asymmetric characteristic.
isBaseFor(y3, y3) because of irreflexive characteristic.

It will correctly infer transitive relation based on SWRL rule given assertions isBaseFor(y1, y2) and isBaseFor(y2, y3).

Thank you for the quick response! It works, I had a mistake in the rule... — António Antunes, Mar 22 '23 at 14:15

"Ignoring transitivity and/or complex subproperty axioms"

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