- 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 That's indeed an outdated one.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