- From: Bernard Vatant <bernard.vatant@mondeca.com>
- Date: Fri, 09 Feb 2007 10:34:24 +0100
- To: "Svensson, Lars" <l.svensson@d-nb.de>
- Cc: public-esw-thes@w3.org
Hi Lars 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 :-) Bernard [1] http://en.wikipedia.org/wiki/Order_relation#Basic_definitions [2] http://en.wikipedia.org/wiki/Congruence_relation#Modular_arithmetic > If so, then related and broader are automatically disjoint since the one is symmetric and the other not. > > Cheers > > Lars (who is preparing to be marked as an illogician) > -- *Bernard Vatant *Knowledge Engineering ---------------------------------------------------- *Mondeca** *3, cité Nollez 75018 Paris France Web: www.mondeca.com <http://www.mondeca.com> ---------------------------------------------------- Tel: +33 (0) 871 488 459 Mail: bernard.vatant@mondeca.com <mailto:bernard.vatant@mondeca.com> Blog: Leçons de Choses <http://mondeca.wordpress.com/>
Received on Friday, 9 February 2007 09:34:32 UTC