- From: Svensson, Lars <l.svensson@d-nb.de>
- Date: Fri, 9 Feb 2007 10:49:46 +0100
- To: "Bernard Vatant" <bernard.vatant@mondeca.com>
- Cc: <public-esw-thes@w3.org>
In litteris suis de Freitag, 9. Februar 2007 10:34, public-esw-thes-request@w3.org <>scripsit: Hi Bernard, > > Good try, but ... >> If we explicitly state that skos:broader is irreflexive and > that related is symmetric, do we really need to explicitly > state that related and broader are disjoint? I might be wrong, > but isn't reflexivity a special case of symmetry (thus "not > reflexive" implies "not symmetric")? > I think you're confusing several things here. Reflexivity and symmetry > are independent properties of relations. > > "Not reflexive" implies "Not symmetric" is obviously wrong. > "neighbourOf" is not reflexive (I'm not my own neighbour), but is > symmetric. > > The other way round reflexivity is not a special case of symmetry. Let > me take an example from arithmetics. > Take "lesser or equal" (<=) relationship or any other order relation > [1], which bears transitivity, reflexivity and antisymmetry, > the latter > meaning > IF (a <= b AND b<=a ) THEN a=b > It is reflexive, but not symmetric > 2 <= 3 but not the other way round. > > It is not necessarily disjoint with some symmetric relationship, like > "congruence modulo 3" [2] 5 ≡ 11 (mod 3) and 5 <= 11 > > Hence the graphs of ≡ (mod 3) and <= are not disjoint > > Quod erat demonstrandum :-) Thanks for the explanation! I think I'll reread my old textbook on logics before posting the next time. Some things are quite different from how you remember them... Cheers Lars (who is now definitely marked as an illogician) -- Dr. Lars G. Svensson Deutsche Nationalbibliothek Informationstechnik Adickesallee 1 60322 Frankfurt http://www.d-nb.de/
Received on Friday, 9 February 2007 09:50:06 UTC