Re: SKOS Comment: skos:broader as subproperty of broaderTransitive seems backwards

On 1 Dec 2008, at 07:50, Booth, David (HP Software - Boston) wrote:

>
> In
> http://www.w3.org/TR/2008/WD-skos-reference-20080829/#L2413
> it says that "skos:broader is not a transitive property", but  
> "skos:broader is a sub-property of skos:broaderTransitive, which is  
> a transitive property".  Isn't this backwards?
>
> By the entailment rules for rdfs:subPropertyOf
> http://www.w3.org/TR/rdf-mt/#rulerdfs7
> if:
>         skos:broader rdfs:subPropertyOf skos:broaderTransitive .
>         uuu skos:broader yyy .
>
> then necessarily:
>
>         uuu skos:broaderTransitive yyy .

David

A subproperty of a transitive property is not necessarily transitive.  
As you rightly point out, if I have

xxx skos:broader yyy

then I can infer

xxx skos:broaderTransitive yyy

However, this doesn't mean skos:broader is transitive. If a relation  
R is transitive, it means that from xRY and yRz, we can infer xRz. So  
the situation with broader is as follows.

If I have

xxx skos:broader yyy
yyy skos:broader zzz

then I can infer (through subproperties) that

xxx skos:broaderTransitive yyy
yyy skos:broaderTransitive zzz

Now, due to the transitivity of broaderTransitive, I can infer

xxx skos:broaderTransitive zzz

This *doesn't* mean that xxx skos:broader zzz, so I don't  
(necessarily) have transitivity of skos:broader.

The WG spent some time discussing this. We believe that there are  
situations where we do not necessarily want skos:broader to be  
transitive, but the pattern we have used allows us to /query/ across  
something which includes the transitive closure of skos:broader,  
without us having to assert that skos:broader is itself transitive.  
We believe that this is the desirable situation.

Hope that helps.

	Sean

--
Sean Bechhofer
School of Computer Science
University of Manchester
sean.bechhofer@manchester.ac.uk
http://www.cs.manchester.ac.uk/people/bechhofer

Received on Monday, 1 December 2008 10:20:42 UTC