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Re: Possible semantic bugs concerning domain and range

From: pat hayes <phayes@ai.uwf.edu>
Date: Fri, 27 Sep 2002 22:37:01 -0500
Message-Id: <p05111b19b9bace8fd999@[65.217.30.172]>
To: Ian Horrocks <horrocks@cs.man.ac.uk>
Cc: www-webont-wg@w3.org

>Pat,
>
>Now we seem to have a come to a better understanding about the
>correspondence between FOL and OWL, could you re-answer the following
>question.
>
>Thanks,
>
>Ian
>
>>Pat,
>>
>>DAML+OIL, and I hope OWL, can be viewed a fragment of FOL, with atomic
>>classes and properties corresponding to unary and binary predicates
>>respectively. According to this correspondence, subClassOf axioms
>>become implications, e.g., A subClassOf B corresponds to:
>>
>>forall x . A(x) -> B(x)
>>
>>Similarly, a property range axiom P range A corresponds to:
>>
>>forall x,y P(x,y) -> A(y).
>>
>>What could be simpler and clearer than that?
>>
>>The combination of these two sentences entails
>>forall x,y P(x,y) -> B(y).
>>
>>What could be simpler and clearer than that?
>  >
>  >If you want some alternative semantics, could you please explain in
>  >similar terms what it is?

Sure. I agree this is clear and simple, and I think everyone agrees 
that something very close to this is what we all want. The issue has 
always been only whether those conditions are necessary, or necessary 
and sufficient. We all want the following to be true:

Range(P, A) -> (forall x,y P(x,y) -> A(y) )

You want

Range(P,A) <-> (forall x,y P(x,y) -> A(y) )

They are about equally clear and intuitive; but the latter rules out 
some possibilities which the former permits. I believe that all the 
'intuitive' entailments that people want in fact hold in both these 
cases; and that the former is therefore to be preferred.

The potential utility of the former is that it allows ranges to have 
properties.  Suppose we wanted to say something about ranges (perhaps 
ranges from a particular class of ranges), expressed by a predicate 
Q:  Range(P, x) -> Q(x), say. (It is SUCH a relief to be able to 
write logic!) With the second, stronger condition, this would entail 
that Q was preserved under implication, ie
(forall x (P(x) -> R(x)) -> (Q(P) -> Q(R))
which is a very strong condition for Q to have to satisfy for no good 
reason; in fact, it is so strong that it would make this practically 
useless, since hardly any useful properties satisfy this kind of 
condition (it is restricted to properties like having more than a 
certain number of instances, things like that.)

I hope this helps to make the point clearer.

Pat

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Received on Friday, 27 September 2002 23:36:45 GMT

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