- From: Richard Cyganiak <richard@cyganiak.de>
- Date: Wed, 16 Jan 2013 18:05:38 +0000
- To: "Patel-Schneider, Peter" <Peter.Patel-Schneider@nuance.com>, Pat Hayes <phayes@ihmc.us>
- Cc: RDF Working Group WG <public-rdf-wg@w3.org>
Peter, Pat, I see -- the ill-typed literal is required to denote something that both is in LV and is not in LV, leading to a contradiction. Going through the ISSUE-109 thread again, I think Antoine proposed a similar mechanism here: http://lists.w3.org/Archives/Public/public-rdf-wg/2012Nov/0194.html Pat didn't like Antoine's approach: http://lists.w3.org/Archives/Public/public-rdf-wg/2012Nov/0196.html Pat proposed what appears to be a different way of achieving the same result: http://lists.w3.org/Archives/Public/public-rdf-wg/2012Nov/0195.html Anyway, I trust that there is *some* way of making this work in the Semantics. Therefore I've re-assigned ISSUE-109 against the RDF Semantics product, assuming that the best way of resolving the issue is to adopt one of the proposals above (plus a small change in Concepts to reflect that semantic change). Best, Richard On 16 Jan 2013, at 16:59, Patel-Schneider, Peter wrote: > During the call today there was some discussion of ill-typed literals. > > *IF* one wants ill-typed literals to be inconsistent then one has to tweak the semantics. > The effect is (roughly) to require that the interpretations for literals whose datatype is in the datatype map belong to the value space for that datatype. This looks a lot like the situation where that literal is range-required to be in the datatype. > > As far as wording goes, RDF semantics would change something like: > > Current: > if <aaa,x> is in D then for any typed literal "sss"^^ddd in V with I(ddd) = x , > if sss is in the lexical space of x then IL("sss"^^ddd) = L2V(x)(sss), otherwise IL("sss"^^ddd) is not in LV > > Revised: > if <aaa,x> is in D then for any typed literal "sss"^^ddd in V with I(ddd) = x , > IL("sss"^^ddd) is in LV > if <aaa,x> is in D then for any typed literal "sss"^^ddd in V with I(ddd) = x , > if sss is in the lexical space of x then IL("sss"^^ddd) = L2V(x)(sss), otherwise IL("sss"^^ddd) is not in LV > > This may look odd, but the net result is that there can be no models for ill-typed literals. > > > Note: I'm not here an advocate for this change, just noting how it could be done. > > peter >
Received on Wednesday, 16 January 2013 18:06:09 UTC