- From: Brian McBride <bwm@hplb.hpl.hp.com>
- Date: Fri, 24 Jan 2003 10:29:59 +0000
- To: marc@jfcarrion.com, marc@jfcarrion.com, www-rdf-comments@w3.org
At 01:27 24/01/2003 -0800, Marc Carrion wrote: [...] > > > Yes and No. Let me explain, and I beg your pardon >again for my english. I understand that any instance >of a class that is subclass of c is instance of c (The >same in UML, Java, C++, ...) but I cannot say in a RDF >instance that (x -rdf:type-> c) (the same in UML, >Java, C++, ...) I'm going to infer it from the >instance and the schema. Does it make sense? Now you want to constrain the abstract syntax so that graphs of the following form x rdf:type c . c rdf:type rdfs:Abstract . Are illegal. If you do that, then you can't make the inference you say you want to make. Maybe if you were to explain what, in practical terms in the system you are building, you need to accomplish, then we could figure something out. For example, lets say you were building a tool that accepted data from a user and its got an adaptive user interface. So it reads a schema file and then uses that to offer the user choices. Now lets say you want to ensure that the user interface does not offer the user the chance to say that something is an instance of one of these union classes you call abstract. You don't need anything in the standard for that. You just define your own notion of an abstract class and go ahead and use it. > I'm not >trying to express a NOT I trying to make a diference >in the range property of the Property rdf:type. Is it >more clear in this way? No. > > > > > If anything identified with a RDF URI it's a > > >Resource, why all the classes I define should > > extend > > >from resource? > > > > I don't know what you mean by "extend from". > Excuse me for my english again (Now that I read that >I don't know why I wrote it) I meant The classes you >define in the schema should be subclass of >rdf:Resource, at least all them are in the schema of >the RDFS. Because everything that RDF can talk about is a resource. Therefore all members of a class must be a resource. Therefore that class must be a subclass of rdfs:Resource. Brian
Received on Friday, 24 January 2003 05:28:49 UTC