- From: Andreas Langegger <al@jku.at>
- Date: Fri, 14 Mar 2008 11:09:34 +0100
- To: Antoine Isaac <Antoine.Isaac@KB.nl>, SKOS <public-esw-thes@w3.org>
- Message-Id: <F1F24EE4-B86A-4E65-891D-E2C40F1EA82D@jku.at>
ah, sure... thanks Antoine. You're right, it's a confusion about
inheritance.
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
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 ;-)
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
Received on Friday, 14 March 2008 10:10:39 UTC