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RE : RE : RE : Suggestion for SKOS FAQ

From: Antoine Isaac <Antoine.Isaac@KB.nl>
Date: Thu, 13 Mar 2008 13:51:05 +0100
Message-ID: <68C22185DB90CA41A5ACBD8E834C5ECD04953D8A@goofy.wpakb.kb.nl>
To: "Sini, Margherita \(KCEW\)" <Margherita.Sini@fao.org>, "Stephen Bounds" <km@bounds.net.au>, "SKOS" <public-esw-thes@w3.org>
Cc: <al@jku.at>
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.

- 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


-------- Message d'origine--------
De: Sini, Margherita (KCEW) [mailto:Margherita.Sini@fao.org]
Date: jeu. 13/03/2008 09:05
: 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:


  A skos:broader B  
  B skos:broader C 
means also:

1)   skos:broader "isA" skos:broaderTransitive which I think is not what we

2) We get the transitivity for free:

  A skos:broaderTransitive B
  B skos:broaderTransitive C
  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:


   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:


   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?

-----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

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.


-------- Message d'origine--------
De: public-esw-thes-request@w3.org de la part de Stephen Bounds
Date: mar. 11/03/2008 22:39
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:


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.


-- 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
>     *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
Received on Thursday, 13 March 2008 12:51:49 UTC

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