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Re: mergeg in current Semantics ED

From: Pat Hayes <phayes@ihmc.us>
Date: Fri, 8 Mar 2013 12:43:04 -0600
Cc: RDF WG <public-rdf-wg@w3.org>
Message-Id: <7379080B-B394-4A2E-8828-0986066DDBDD@ihmc.us>
To: Peter Patel-Schneider <pfpschneider@gmail.com>, Antoine Zimmermann <antoine.zimmermann@emse.fr>

Ah. Yes, now I see your (and Antoine's) point. Antoine, sorry I was slow. 

I think we need a new idea to handle this. 

Observation: a bnode scope might extend over several graphs, but a graph cannot cross bnode scopes. So bnode scopes are a 'larger' syntactic unit than graphs. Given a bnode scope, the set of triples contained in it is an RDF graph, call it the /scope graph/. Every RDF graph is a subgraph of some scope graph. 

A graph is /complete/ (/saturated/? /scoped/? /whole/? /coherent/? /molecular/?) when, if it contains a bnode, then it contains all triples in the scope graph which contain that bnode. That is, for each bnode b in the scope, it either does not contain b, or it contains every triple containing b. 

Ground graphs and scope graphs are always complete. A union (= merge) of complete graphs is complete.  Any set of complete graphs entails its union. Antoine's example shows that this may not be true for incomplete graphs. 

Comments, before I start editing?


On Mar 7, 2013, at 4:56 PM, Peter Patel-Schneider wrote:

> On Thu, Mar 7, 2013 at 2:41 PM, Pat Hayes <phayes@ihmc.us> wrote:
> On Mar 7, 2013, at 3:12 PM, Peter Patel-Schneider wrote:
> > I think that the current document makes the entailment not work.
> >
> > G1 is Ex p1(s1,x)
> > G2 is Ex p2(s2,x)
> No, its not. If they share a blank node, they must be in  the same scope; and the existential is defined at the scope, not at the graph, level. So under the conditions given by Antoine, the Ex is outside the conjunction of G1 and G2.
> I see exactly the opposite of this in the current document.
> "If E is an RDF graph then I(E) = true if [I+A](E) = true for some mapping A from the set of blank nodes in the scope of E to IR, otherwise I(E)= false."
> So, consider I with I(s1) = s1, I(s2) = s2, I(p1) = p1, I(p2) = p2, IEXT(p1) = {<s1,s1>}, IEXT(p2) = {<s2,s2>}
> (add the other stuff to minimally turn this into a simple interpretation).
> Then I(G1) = true, from the mapping A1:x->s1
> and I(G2) =true, from the mapping A2:x->s2
> but there is no mapping for x that makes I({G1,G2}) = true.
> So, yes, the existential in {G1,G2} is global in the current document, but that is precisely what makes the difference.
> peter

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Received on Friday, 8 March 2013 18:43:45 UTC

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