Re: Challenge for RDF Gurus :)

Seth Russell writes:

> From: <tarod@softhome.net>
> 
> > > re: http://robustai.net/mentography/rdfs_domain_range2.gif
> > >
> > >
> > > >   Good try but I must say that it's not 100% what I asked for because
> for
> > > > the range issue you use
> > > >   Class C
> > > >   A is subClassOf C
> > > >   B is subClassOf C
> > > >   And then c range is C. It's a good aproach but it's not logically
> > > > correct, you are saying that range of c is (C or A or B) and I asked
> for
> > > > range of c should be (A or B)
> > >
> > > Ok, I saw this problem after I published the graph.  I would need a way
> to
> > > say that there is no instances of C which is not and instance of A or B.
> > > I'm beginning to agree with Sean, there is no way to say this with the
> > > primitives of rdfs only.
> >   It was posible before some RDFCore changes :)
> >
> > > What is your objection to using the daml schema?
> >
> >   I have no objection, this is just a challenge.
> >
> > > >   Now try it with the old aproach it's easier.
> > >
> > > What approach are you talking about here?
> >
> >   Before some changes in the schema, that a property had two domains (at
> > the begining a property must only have one range, now it can have more
> than
> > one) means that the subject of the property must be in one of those
> > domains, it was a disjuntion of restrictions. When they added more than
> one
> > range if they had used this vision, the value of a property must be a
> > member of one of the domains, if that make sense to you, try it now. It's
> > very easy having this in mind.
> 
> Well if the domain restraint is jisunctive and the range restraint is
> conjunctive, then I suppose your example would eaisly work that old way.
> But if one wanted the opposite case ( range (A and B).  domain ( A or B)),
> then we still couldn't do it.   Intiitively don't we want domain and range
> to be symmetric here?
> 
> Mentograph available upon request.
> Seth Russell
> 
> 

  Sorry, I meant both of them should be disjuntive.

      Marc