*"I don't think so. I think it is much simpler than that. If I have understood it correctly, it is a consequence of "lean"ing the graph."* Not sure if I understand, David, doesn't that imply you're simply doing a structural equivalence check, defining a composite key consisting of all properties when it's possible not all properties are required? Anthony On Tue, Dec 4, 2018 at 8:11 PM David Booth <david@dbooth.org> wrote: > >> On Tue, Dec 4, 2018 at 5:42 PM Patrick J Hayes wrote: > >> How can you ever know that the particular arguments given to > >> the n-ary relation are enough to constitute a key? > > I'm not sure that "key" is quite the right term here, though > perhaps it amounts to a key of sorts, because the connected > properties uniquely "identify" a lean version of that blank > node. I think. I'm still trying to get my head around this! > > >> And if you do know it, how can you state it so as to validate > >> the inference? I don't think (?) it can be stated in, for > >> example, OWL. You would need at least FOL with identity, > >> way beyond RDF expressivity. > >> > > On 12/4/18 8:51 PM, Anthony Moretti wrote: > > It think it's the same as the discussion around Addresses, > > it just depends from type to type what constitutes a key. > > I don't think so. I think it is much simpler than that. If I > have understood it correctly, it is a consequence of > "lean"ing the graph. > > David Booth > >Received on Wednesday, 5 December 2018 17:24:05 UTC
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