Re: Dark triples, current closure / entailment rules, can someone clarify?

At 06:59 PM 6/18/02 +0100, Jan Grant wrote:

>Pat indicated at the F2F that entailments "accidentally" drawn would be
>"harmless". Can someone clear up this example for me?
>
>A:
>         <eg:foo1> <rdf:subPropertyOf> <dark:eg:foo2> .
>         <dark:eg:foo2> <rdf:subProperotyOf> <eg:foo3> .
>
>         <eg:a> <eg:foo1> <eg:b> .
>
>B:
>         <eg:foo1> <rdf:subPropertyOf> <dark:eg:foo2> .
>         <dark:eg:foo2> <rdf:subProperotyOf> <eg:foo3> .
>
>         <eg:a> <dark:eg:foo2> <eg:b> .
>
>C:
>         <eg:foo1> <rdf:subPropertyOf> <dark:eg:foo2> .
>         <dark:eg:foo2> <rdf:subProperotyOf> <eg:foo3> .
>
>         <eg:a> <eg:foo3> <eg:b> .
>
>
>Does A |= C? Does A |= B? B |= C?
>
>Awaitening en(light|dark)enment,
>jan

My opinions:

A |= B:  yes.

B |= C:  no.   (e.g. consider interpretation in which all relational 
extensions are empty.)

A |= C:  Hmmm.  I think the answer is yes.  I think the model theory 
constrains the relational extensions of eg:foo1, dark:eg:foo2 and eg:foo3 
to have a subset relationship, even though the relational extension of 
dark:eg:foo2 does not affect the truth of any graph that uses it.  Which 
leads to the interesting observation that entailment is not transitive, 
which I inituitively expect it to be.

Good question!

I don't remember the context of accidental entailments mentioned, so cannot 
comment further.

#g


-------------------
Graham Klyne
<GK@NineByNine.org>