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RE: SKOS Comment: skos:broader as subproperty of broaderTransitive seems backwards

From: Booth, David (HP Software - Boston) <dbooth@hp.com>
Date: Mon, 1 Dec 2008 13:46:47 +0000
To: Sean Bechhofer <sean.bechhofer@manchester.ac.uk>
CC: "public-swd-wg@w3.org" <public-swd-wg@w3.org>
Message-ID: <CD2B872281385A439B98164F5351E6DD39C4779294@GVW1144EXB.americas.hpqcorp.net>

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 13:48:10 UTC

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