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Re: builtins operating on rif:iri

From: Sandro Hawke <sandro@w3.org>
Date: Fri, 24 Apr 2009 07:12:05 -0400
To: Jos de Bruijn <debruijn@inf.unibz.it>
cc: kifer@cs.sunysb.edu, public-rif-wg@w3.org
Message-ID: <23838.1240571525@ubehebe>

...
> > ... and since it's not the same in every interpretation, it's not
> > entailed.   I think I'm getting that.  So...
> > 
> > positive-entailment
> > PREMISE:    (empty)
> > CONCLUSION: member(eg:foo, list(eg:foo eg:bar))
> > 
> > negative-entailment
> > PREMISE:    (empty)
> > CONCLUSION: member(eg:baz, list(eg:foo eg:bar))
> > 
> > positive-entailment
> > PREMISE:    eg:baz = eg:bar
> > CONCLUSION: member(eg:baz, list(eg:foo eg:bar))
> > 
> > Agreed?
> 
> agreed.
...
> > 
> > positive-entailment
> > PREMISE:    eg:bar = 1
> >             eg:foo = 2
> > CONCLUSION: index-of(list(eg:foo eg:bar), eg:foo) = list(1)
> 
> yes
> 
> > 
> > positive-entailment
> > PREMISE:    eg:bar = eg:foo
> > CONCLUSION: index-of(list(eg:foo eg:bar), eg:foo) = list(1 1)
> 
> Did you mean List(1 2)?

Yes, sorry.  (And I suspect we really want 0-indexed lists, but I'll
stick with 1-indexed lists for this conversation, since I accidentally
started that way.)

> > But what if the premise is empty?  I'm thinking that none of these
> > entailments would hold...  (so these would be valid tests)....
> > 
> > negative-entailment
> > PREMISE:    
> > CONCLUSION: index-of(list(eg:foo eg:bar), eg:foo) = list(1 1)
> 
> Yes
> 
> > 
> > negative-entailment
> > PREMISE:    
> > CONCLUSION: index-of(list(eg:foo eg:bar), eg:foo) = list(1)
> 
> No. eg:foo is mapped to the same object in every interpretation. Let's
> call this object a. And let's say eg:foo is mapped to b. Then
> list(eg:foo eg:bar) represents the sequence (a,b). And certainly the
> object a appears in the first position of this sequence.
> 
> So, this should be a positive entailment test.

Well, I'd agree that in all interpretations, there's a match in the
first position.  But in some interpretations there's also a match in the
second postion (namely interpretations where eg:foo=eg:bar).  So, in
some interpretations, the left side of the equals is interpretated as
list(1) and in others it's list(1 2).  So, I think that means the
equality doesn't hold in all interpretations, and the proposed
conclusion is not entailed...

Right?

     -- Sandro
Received on Friday, 24 April 2009 11:12:18 GMT

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