- From: Antoine Isaac <Antoine.Isaac@KB.nl>
- Date: Wed, 12 Mar 2008 13:09:00 +0100
- To: "Stephen Bounds" <km@bounds.net.au>, "SKOS" <public-esw-thes@w3.org>
- Cc: "Sini, Margherita \(KCEW\)" <Margherita.Sini@fao.org>, <al@jku.at>
- Message-ID: <68C22185DB90CA41A5ACBD8E834C5ECD04953D7E@goofy.wpakb.kb.nl>
Thanks a lot Stephen for your clarification. I would actually add: at some point we considered in the WG (and I was supporting this) that broaderTransitive could be actually a subproperty of skos:broader. This actually would have matched cases for which you allow skos:broader to be locally transitive (that is, on certain KOSs and not on others), which is what we wanted (and still allow, on the condition that KOS creators explicitly assert the 'extra' A skos:broader C -kind of links). But this was judged less convenient. Because then if you want to say that the broaders of a given KOS are transitive, you have to *explicitly assert* statements of broaderTransitive. While with the current version, you get the transitivity for free: whenever you assert a broader, there is a transitive one that is inferred for it, de facto building a transitive hierarchy for your KOS. Meanwhile, you can still access your original skos:broader statements, without having them messed up by the transitivity. Antoine -------- Message d'origine-------- De: public-esw-thes-request@w3.org de la part de Stephen Bounds Date: mar. 11/03/2008 22:39 À: SKOS Cc: Sini, Margherita (KCEW); al@jku.at Objet : Re: RE : Suggestion for SKOS FAQ Hi Margaret & Andy, I thought that too when I first looked at the SKOS Primer, but you need to remember that OWL sub-properties are subtractive, not additive. Another way of putting this is that super-properties make *less* restrictive statements about the world. The full hierarchy of skos:broader is: skos:semanticRelation skos:broaderTransitive skos:broader Which means that for A skos:broader B, this entails that: A skos:broaderTransitive B and A skos:semanticRelation B We can't reverse the order of skos:broaderTransitive and skos:broader in the because of the transitive case. If: A skos:broaderTransitive B and B skos:broaderTransitive C then A skos:broaderTransitive C but A skos:broader C is NOT entailed If skos:broader were a super-property of skos:broaderTransitive, this statement would also need to be true. Regards, -- Stephen. Sini, Margherita (KCEW) wrote: > I agree with Andy, I also think it should be a sub-property, not a > super-property... > > Regards > Margherita > > -----Original Message----- > *From:* public-esw-thes-request@w3.org > [mailto:public-esw-thes-request@w3.org] *On Behalf Of *Andreas Langegger > *Sent:* 11 March 2008 12:14 > *To:* Alasdair J G Gray > *Cc:* Antoine Isaac; Simon Spero; iperez@babel.ls.fi.upm.es; SKOS > *Subject:* Re: RE : Suggestion for SKOS FAQ > > Hi, > > first I din't pay much attention to your discussion, because I > thought this case is clear... looking at the spec I read > "skos:broaderTransitive owl:subClassOf skos:broader" - but there it > says (to my surprise): skos:broaderTransitive and others are "super > properties" - why that? > > If I would model this I would say: > > skos:semanticRelation a owl:ObjectProperty . > skos:broader a skos:semanticRelation . > skos:narrower a skos:semanticRelation . > skos:broaderTransitive a skos:broader; a owl:TransitiveProperty . > skos:narrowerTrasnsitive a skos:narrower; a owl:TransitiveProperty . > and so on... > > can anybody comment on this why the specs says "super property" and > not "sub property" ? > Whith the statements above I can deceide whether to allow > transitivity or not. And because of OWA, skos:broader not explicitly > asserted as a transtive property, it does not mean, that it _cannot > be_ transitive, sure it can, but it does not need to be valid. > > If a taxonomy should be ISO2788 compliant, just use the *Transitive > versions - so it's up to the modeler and not to the application > which I think is fine. > > regards > Andy
Received on Wednesday, 12 March 2008 12:09:17 UTC