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RE: RE : RE : RE : RE : Suggestion for SKOS FAQ

From: Sini, Margherita (KCEW) <Margherita.Sini@fao.org>
Date: Tue, 18 Mar 2008 17:55:29 +0100
To: Antoine Isaac <Antoine.Isaac@KB.nl>, al@jku.at, SKOS <public-esw-thes@w3.org>
Message-id: <BA453B6B6B217B4D95AF12DBA0BFB669029DB037@hqgiex01.fao.org>
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	That's indeed an outdated one.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/> 
	
	
	

Received on Tuesday, 18 March 2008 16:56:25 GMT

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