- From: Antoine Isaac <aisaac@few.vu.nl>
- Date: Mon, 01 Dec 2008 14:53:54 +0100
- To: "Booth, David (HP Software - Boston)" <dbooth@hp.com>
- CC: "public-swd-wg@w3.org" <public-swd-wg@w3.org>
Hi David, There is a small note in the SKOS Primer about this issue [1]. Your feedback on whether this is appropriate would be warmly welcome! Best regards, Antoine Isaac [1] http://www.w3.org/TR/2008/WD-skos-primer-20080829/#sechierarchy > Hi Sean, > > Good point and explanation! So skos:broader does have *some* transitivity, but it is only transitive to the extent that a skos:broader relations is a skos:broaderTransitive relation -- the semantics of skos:broader that are distinct from the semantics of skos:broaderTranstive are not transitive. > > It might be worth adding or linking to your explanation in a future draft, since others may puzzle over this also. > > Thanks > > David Booth, Ph.D. > HP Software > +1 617 629 8881 office | dbooth@hp.com > http://www.hp.com/go/software > > Statements made herein represent the views of the author and do not necessarily represent the official views of HP unless explicitly so stated. > > >> -----Original Message----- >> From: Sean Bechhofer [mailto:sean.bechhofer@manchester.ac.uk] >> Sent: Monday, December 01, 2008 5:18 AM >> To: Booth, David (HP Software - Boston) >> Cc: public-swd-wg@w3.org >> Subject: Re: SKOS Comment: skos:broader as subproperty of >> broaderTransitive seems backwards >> >> >> On 1 Dec 2008, at 07:50, Booth, David (HP Software - Boston) wrote: >> >>> In >>> http://www.w3.org/TR/2008/WD-skos-reference-20080829/#L2413 >>> it says that "skos:broader is not a transitive property", but >>> "skos:broader is a sub-property of skos:broaderTransitive, which is >>> a transitive property". Isn't this backwards? >>> >>> By the entailment rules for rdfs:subPropertyOf >>> http://www.w3.org/TR/rdf-mt/#rulerdfs7 >>> if: >>> skos:broader rdfs:subPropertyOf skos:broaderTransitive . >>> uuu skos:broader yyy . >>> >>> then necessarily: >>> >>> uuu skos:broaderTransitive yyy . >> David >> >> A subproperty of a transitive property is not necessarily transitive. >> As you rightly point out, if I have >> >> xxx skos:broader yyy >> >> then I can infer >> >> xxx skos:broaderTransitive yyy >> >> However, this doesn't mean skos:broader is transitive. If a relation >> R is transitive, it means that from xRY and yRz, we can infer xRz. So >> the situation with broader is as follows. >> >> If I have >> >> xxx skos:broader yyy >> yyy skos:broader zzz >> >> then I can infer (through subproperties) that >> >> xxx skos:broaderTransitive yyy >> yyy skos:broaderTransitive zzz >> >> Now, due to the transitivity of broaderTransitive, I can infer >> >> xxx skos:broaderTransitive zzz >> >> This *doesn't* mean that xxx skos:broader zzz, so I don't >> (necessarily) have transitivity of skos:broader. >> >> The WG spent some time discussing this. We believe that there are >> situations where we do not necessarily want skos:broader to be >> transitive, but the pattern we have used allows us to /query/ across >> something which includes the transitive closure of skos:broader, >> without us having to assert that skos:broader is itself transitive. >> We believe that this is the desirable situation. >> >> Hope that helps. >> >> Sean >> >> -- >> Sean Bechhofer >> School of Computer Science >> University of Manchester >> sean.bechhofer@manchester.ac.uk >> http://www.cs.manchester.ac.uk/people/bechhofer >> >> >> >> > >
Received on Monday, 1 December 2008 13:54:32 UTC