- From: Yuzhong Qu <yzqu@seu.edu.cn>
- Date: Tue, 3 Aug 2004 17:22:20 +0800
- To: <ecky@free.fr>
- Cc: "RDF Interest" <www-rdf-interest@w3.org>
> >Hi, All > > > >At the begining of the second paragraph in section 4.4 RDFS > >Entailment , it says > > "Since every > ><http://www.w3.org/TR/rdf-mt/#rdfsinterpdef>rdfs-interpretation is > >an <http://www.w3.org/TR/rdf-mt/#rdfinterpdef>rdf-interpretation, if > >S **rdfs-entails** E then it **rdf-entails** E;". > > > >Maybe, it should be changed to: > > > >..., if S rdf-entail E then it rdfs-entails E;... > > > >Is it correct? > > > > Yes, you are correct. > *Are you sure ? If rdf-entail is more general than rdfs-entail and both of them represent some sort of implication, it is just the way like it is in the document and not the other way round. Yes. Let's prove the following proposition: if S rdf-entails E then S rdfs-entails E. Suppose S rdf-entails E. Let I be a given rdfs-interpretation satifying S. We have that I satifies E due to the following facts: 1. I is also a rdf-interpretation ( Remember that every rdfs-interpretation is an rdf-interpretation) 2. I satisfies S 3. S rdf-entails E So, E is satisfied by any given rdfs-interpretation satifying S. It means S rdfs-entails E. This is basic in general Model Theory. Yuzhong Qu
Received on Tuesday, 3 August 2004 05:18:38 UTC