- From: Andreas Langegger <al@jku.at>
- Date: Tue, 18 Mar 2008 22:58:34 +0100
- To: "Sini, Margherita (KCEW)" <Margherita.Sini@fao.org>
- Cc: Antoine Isaac <Antoine.Isaac@KB.nl>, SKOS <public-esw-thes@w3.org>
yes. I think this is not possible, because the application or user can
always select triples or execute queries referring to
skos:transitiveBroader which will result in the transitivity (this one
which is ment "for free" ;-) but in some cases may be forbidden)
so, that's why skos:broaderIntransitive was suggested earlier in this
thread I guess...
regards
Andy
On Mar 18, 2008, at 5:55 PM, Sini, Margherita (KCEW) wrote:
> Dear all,
>
> I understood the point explained by Stella and I agree.
>
> But i would like to return to the proposal to make
> skos:broaderTransitive a
> super-property of skos:broader.
>
> I understood the point of Alistar that it is possible to have non-
> transitive
> sub-properties of transitive properties and viceversa. I agree on
> that. I
> also clarify on my mind inheritance.
>
> But: what if i would like to express:
> A skos:broader B
> B skos:broader C
> and assert that A not-skos:broader C?
>
> in other words i would like to express intransitivity. How do i do
> that?
>
> Also: is this topic associated to some issue? maybe issue 44 or
> there is a
> more specific one?
>
> Thanks
> Margherita
>
> -----Original Message-----
> From: public-esw-thes-request@w3.org on behalf of Antoine Isaac
> Sent: Fri 3/14/2008 11:26
> To: al@jku.at; SKOS
> Cc:
> Subject: RE : RE : RE : RE : Suggestion for SKOS FAQ
>
>
>
>
> Dear Andy,
>
> >> Is there a current draft for the new SKOS core as rdf? I can only
> find
> this one which doesn't include the new transitivity solution:
> http://www.w3.org/2004/02/skos/core.rdf
> >>
>
> That's indeed an outdated one. There should be a new one by the time
> SKOS goes candidate recommendation, but for the moment there is
> nothing
> available
>
> >>
> I assume skos:transitiveBroader is an owl:TransitiveProperty
> (http://www.w3.org/TR/owl-ref/#TransitiveProperty-def
> ). If skos:broader is a sub property of skos:broaderTransitive this
> does not imply skos:broader is also an owl:TransitiveProperty! - it
> is
> no sub class it's just a sub property ;-)
> So, if I use the skos:whateverTransitive property e.g. in a SPARQL
> query, I get the whole transitive closure of the concept relation and
> if I use skos:broader/etc. I only get direct assertions, right? With
> this approach it's possible to interpret relations as transitive by a
> query/application although the author of the KOS did not even use
> transitive properties, right? That's fine ;-)
> >>
>
> That's *exactly* this!
>
> Cheers,
>
> Antoine
>
> Thanks,
> Andy
>
> On Mar 13, 2008, at 1:51 PM, Antoine Isaac wrote:
>
> > I'm sorry I don't have time to read all your mail and answer point
> > by point.
> >
> > But it seems really related to confusion about transitivity and
> > inheritance.
> > You indeed assume that if something is transitive, then it has more
>> information defined, and thus should be a sub-property of a non-
> > transitive property.
> >
> > But it is prefectly possible to say that a superproperty is
> > transitive. That says something about its graph (that is, the
> > couples (x,y) that are related by the property, as the As and Bs in
>> your example).
> > But now, if you have a sub-property, formally it is defined as a
> sub-
> > part of the graph. So you lose elements (couples), and something
> > that was true at the level of the super-property (e.g.
> transitivity)
> > might not be true anymore for the sub-property.
> >
> > Example:
> > - one property 'blob1' defined by the graph {(a,b), (b,c), (a,c)}
> > (it relates only these elements) is transitive
> > - one property 'blob2' defined by the graph {(a,b), (b,c)}
> > blob2 is a sub-property of blob1 and yet blob2 is not transitive.
> >
> > That's what happens currently in SKOS, where blob1 is
> > broaderTransitive and blob2 is broader
> >
> > Antoine
> >
> >
> > -------- Message d'origine--------
> > De: Sini, Margherita (KCEW) [mailto:Margherita.Sini@fao.org]
> > Date: jeu. 13/03/2008 09:05
> > À: Antoine Isaac; Stephen Bounds; SKOS
> > Cc: al@jku.at
> > Objet : RE: RE : RE : Suggestion for SKOS FAQ
> >
> > Dear all,
> >
> > I appreciate the efforts from Alistar, Stephen and Simon to explain
>> this, but
> > (sorry) unfortunately I am not convinced.. maybe I miss
> something....
> > Let me summarize from my point of view so that you can tell me if I
>> am wrong:
> >
> > - we say that for skos:broader we could not say if it is transitive
> or
> > intransitive (it may be or not be = could be locally transitive but
>> could be
> > also not transitive).
> > - we say that if somebody want to say that they broader
> > relationships is
> > really transitive, can use a specific one "broaderTransitive"
> > - I think that in OWL, when we say subclassof we actually means "is
> A"
> >
> > So this situation:
> >
> > skos:semanticRelation
> > skos:broaderTransitive
> > skos:broader
> >
> > A skos:broader B
> > B skos:broader C
> >
> > means also:
> >
> > 1) skos:broader "isA" skos:broaderTransitive which I think is not
>> what we
> > want...
> >
> > 2) We get the transitivity for free:
> >
> > A skos:broaderTransitive B
> > B skos:broaderTransitive C
> > therefore
> > A skos:broaderTransitive C
> >
> > ... But what about if I wanted to say that
> >
> > A skos:broader B
> > B skos:broader C
> >
> > and they are not transitive?
> >
> > I think that if somebody wanted the trasitivity NEEDED to
> > *explicitly assert*
> > statements ... otherwise we assume that all skos:broader are also
> > skos:broaderTransitive, no?
> >
> > Then we have:
> >
> > - super-properties make *less* restrictive statements about the
> world.
> > skos:broader I think is less restrictive than
> skos:broaderTransitive
> > because
> > as I understood "skos:broader" we not not know about Transitivity,
> but
> > "skos:broaderTransitive" IS transitive, so IT is more restrictive,
> no?
> >
> > Then we have:
> >
> > >>>We can't reverse the order of skos:broaderTransitive and
> > skos:broader in
> > the because of the transitive case. If:
> > <<<
> >
> > skos:semanticRelation
> > skos:broader
> > skos:broaderTransitive
> >
> > A skos:broaderTransitive B and
> > B skos:broaderTransitive C then
> > A skos:broaderTransitive C but
> >
> > A skos:broader C YES because in this case we agreed that A and
>> B and B
> > and C are related by transitite broader
> >
> >
> > Therefore I can propose another solution:
> >
> > skos:semanticRelation
> > skos:broader
> > skos:broaderTransitive
> > skos:broaderIntransitive
> >
> > A skos:broaderTransitive B and
> > B skos:broaderTransitive C then
> > A skos:broaderTransitive C but
> > therefore A skos:broader C --> is correct to arrive here
> >
> > A1 skos:broaderIntransitive B1 and
> > B1 skos:broaderIntransitive C1 then
> > A1 and C1 are not related
> > therefore we cannot say A1 skos:broader C1 which is correct to
>> arrive to
> > this conclusion because we agreed that A1 is broader than B1 and b1
>> broader
> > than C1 but in an intransitivity way...
> >
> > Where I am wrong?
> > Thanks
> > Margherita
> >
> >
> > -----Original Message-----
> > From: Antoine Isaac [mailto:Antoine.Isaac@KB.nl]
> > Sent: 12 March 2008 13:09
> > To: Stephen Bounds; SKOS
> > Cc: Sini, Margherita (KCEW); al@jku.at
> > Subject: RE : RE : Suggestion for SKOS FAQ
> >
> >
> > 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
> >
> >
>
>
>
> ----------------------------------------------------------------------
> Dipl.-Ing.(FH) Andreas Langegger
> Institute for Applied Knowledge Processing
> Johannes Kepler University Linz
> A-4040 Linz, Altenberger Straße 69
> http://www.langegger.at <http://www.langegger.at/>
>
>
>
>
----------------------------------------------------------------------
Dipl.-Ing.(FH) Andreas Langegger
Institute for Applied Knowledge Processing
Johannes Kepler University Linz
A-4040 Linz, Altenberger Straße 69
http://www.langegger.at
Received on Tuesday, 18 March 2008 21:59:53 UTC