# Re: AW: [SKOS] semantics

From: Bernard Vatant <bernard.vatant@mondeca.com>
Date: Fri, 09 Feb 2007 10:34:24 +0100
Message-ID: <45CC4020.1090403@mondeca.com>
To: "Svensson, Lars" <l.svensson@d-nb.de>

```
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
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```
Received on Friday, 9 February 2007 09:34:32 GMT

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