From: (unknown charset) 梅婧 <mayyam@is.pku.edu.cn>

Date: Tue, 09 Nov 2004 17:04:59 +0800

To: (unknown charset) www-rdf-interest@w3.org

Message-ID: <IsMailClient-F200411091704.AA04590015@is.pku.edu.cn>

Date: Tue, 09 Nov 2004 17:04:59 +0800

To: (unknown charset) www-rdf-interest@w3.org

Message-ID: <IsMailClient-F200411091704.AA04590015@is.pku.edu.cn>

hello, as it was mentioned that RDF semantics is monotonic, i.e., Suppose S is a subgraph of S' and S entails E. Then S' entails E. however, according to http://www.w3.org/TR/owl-absyn/rdfs.html (OWL Web Ontology Language Semantics and Abstract Syntax Section 5. RDF-Compatible Model-Theoretic Semantics), the If-and-only-if conditions is too strong to destroy the monotonicity !!! for example, it was said that "if c \in CEXT(S(owl:SymmetricProperty)) iff <x,y> \in EXT(c) implies <y,x> \in EXT(c)" in short, c is a SymmetricProperty iff c(x,y)->c(y,x). ok, now we turn to the RDF graph. firstly, we only have two facts "c(x,y), c(y,x)" to be the set S={c(x,y), c(y,x)}. then, S entail E={c is a SymmetricProperty} (what is "if" tells us) secondly, we add a new fact "c(x,z)" to have the set S'={c(x,y), c(y,x), c(x,z)}. of course, c is not a SymmetricProperty any more, at least it is unknown now! i.e., S' cannot entail E! Sorry for bothering you all, however i wonder if i am right or just i have done some stupid mistakes? Best wishes, Sincerely your, Jing MeiReceived on Tuesday, 9 November 2004 09:10:44 UTC

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