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Re: [TF:DbE] The easiest keys there are

From: Bijan Parsia <bparsia@cs.man.ac.uk>
Date: Wed, 3 Oct 2007 19:22:10 +0100
Message-Id: <E2E6C7F2-3556-4AE7-A5B8-0785E4336638@cs.man.ac.uk>
Cc: ewallace@cme.nist.gov, public-owl-dev@w3.org
To: Matthew Pocock <matthew.pocock@ncl.ac.uk>

On 3 Oct 2007, at 18:12, Matthew Pocock wrote:

> On Wednesday 03 October 2007, Bijan Parsia wrote:
>> Quick check reveals to me that you got why more elaborate key
>> reasoning is hard (i.e., why we don't want to have to work with
>> unnamed individuals or unknown key values and restrictions on key
>> properties). I took Evan to be asking about why we can't have "check"
>> semantics for keys (i.e., missing keys are a violation).
> Ah, my bad.

No worries.

> So - is the "check" semantics equivalent to raising "inconsistent"
> if the values of the key for all relevant individuals are not  
> entailed by the
> ontology?

No. At least, I don't think so. Here, the check is *whether* you have  
a *known* key. The kind of thing you can express with the K and A  

> That is, given an ontology o, some individual i, a key k, and a
> value v (all in the interpretation of o),
> i in (k some Thing) => exists v s.t. o |= k(i, v)

Oh, I see. The issue would be that it has to be the *same* v in all  
models. I.e., if i k (4 or 5), then there's one model where it's 4  
and one where it's 5. So it has *a* key in all models, but you don't  
know what it is. (I'm confused whether v is a variable or a constant  
in your example, given the quantifier.) Even this is a little  
tendentious since I'm sorta working off an MKNFy account. When you  
have the ground abox statement i k v, obviously, in every model k  
contains the pair (i,v).

Received on Wednesday, 3 October 2007 18:21:15 UTC

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