- From: Benja Fallenstein <b.fallenstein@gmx.de>
- Date: Fri, 14 May 2004 00:09:37 +0300
- To: "Burkett, Bill" <WBurkett@modulant.com>
- Cc: www-rdf-interest@w3.org
-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 Burkett, Bill wrote: | The problem I have is an apparent contradiction in the latter two | statements with respect to m. As a member of C, members of M are | *not* also members of C. If x is a member of X, which is a member of C, this does not imply that x is a member of C. However, it doesn't preclude it either. In itself, it doesn't say anything about whether x is a member of C or not. Both is possible. It's simply an unrelated statement: ~ a member b; b member c says (in RDF) nothing about whether a is a member of c or not. You seem to imply that from "m member M; M member C" it follows that "m not member C." This is *not* correct. If it were, then there would be a paradox, but it isn't so. | As a subclass of C, members of M *are* also | members of C. This is correct. | Is this where we find the Incompleteness of our | rdf/rdfs representational langauge? Is this an inherent | paradox/contradiction that we just have to live with? I don't think there's a paradox or contradiction on the formal level. | You've stressed the importance of a "metaclass", Benja. Is there some | aspect of this term that I'm failing to understand in my understanding | of this situation? I don't think that's the problem -- I think the problem is that you think that "looking at it from a certain angle," ~ a rdf:type b ~ b rdf:type c seems to you to imply that ~ a NOT rdf:type c However, that implication is simply not true, looking at it from any angle. The first two statements are simply independent of the third -- as indepent as ~ A foaf:knows B is from ~ A foaf:familyName "Griffin" Your "As a member of C, members of M are *not* also members of C" is along the level of "As someone who knows B, A does not have the family name 'Griffin'" -- it's correct when you interpret it in a kind of weird way, but it's not correct if interpreted as "If A knows B, then A cannot have 'Griffin' as their family name." Does this help? - - Benja -----BEGIN PGP SIGNATURE----- Version: GnuPG v1.2.4 (GNU/Linux) Comment: Using GnuPG with Thunderbird - http://enigmail.mozdev.org iD8DBQFAo+QAUvR5J6wSKPMRAvlvAKDCeZAHdRj/HQTlAe6qlunzfmB/PQCfeENK JslPTRVG1Y3V7dFHXmVU0MQ= =EMEz -----END PGP SIGNATURE-----
Received on Thursday, 13 May 2004 17:10:10 UTC