Re: Tests rdfms-seq-representation-test00[24]

On Fri, 7 Nov 2003, Jeremy Carroll wrote:

>
>
> These tests are incorrect.
>
> They are failed by Jena and 3Store and passed by Surnia and Euler.
>
> The tests are
>
> *empty*
>
> entails
>
> rdf:_1 rdf:type rdfs:ContainerMembershipProperty .
>
> and
>
> rdf:_1 rdfs:subPropertyOf rdfs:member .
>
>
> The empty interpretation (i.e. an interpetation of the empty vocabulary)
> provides a counterexample. (With true implies false).
>
> We have an outstanding comment from Dave Reynolds from 25 July to this
> effect.
>
> Suggest change the premises to
>
> eg:foo rdf:_1 "bar" .
>
> Jeremy

These tests appear to be true according to the latest copy of the
editor's draft: that is, the "core RDF vocabulary" subset of rdfV has
been dropped. The appropriate prose from the ed's draft now is:

[[
An rdfs-interpretation of V is an rdf-interpretation I of (V union rdfV
union rdfsV) which satisfies the following semantic conditions and all
the triples in the subsequent table, called the RDFS axiomatic triples,
which contain only names from (V union rdfV union rdfsV).
]]

This reflects, I think, the abandonment of the attempt to shoehorn
forward-reasoner rules into the semantics (Pat?) so the test case is
invalid if one reads the LC draft but is valid according to the ed's
draft.

If the ed's draft represents a final position on this then the test case
is accurate (although it is impossible for a last-call reviewer to
observe this).


-- 
jan grant, ILRT, University of Bristol. http://www.ilrt.bris.ac.uk/
Tel +44(0)117 9287088 Fax +44 (0)117 9287112 http://ioctl.org/jan/
You know something's gone badly wrong when your algorithm
takes O(n^2) time but uses O(2^n) space.

Received on Friday, 7 November 2003 12:36:37 UTC