From: Fred Zemke <fred.zemke@oracle.com>

Date: Thu, 12 Jan 2006 09:31:59 -0800

Message-ID: <43C6928F.2090708@oracle.com>

To: public-rdf-dawg-comments@w3.org

Date: Thu, 12 Jan 2006 09:31:59 -0800

Message-ID: <43C6928F.2090708@oracle.com>

To: public-rdf-dawg-comments@w3.org

5.4 Optional matching - formal definition The last paragraph in the box does not make sense. Reducing the sentence to its logical connectives, its form is "If P then Q if R otherwise if S", which evidently should be grouped "If P then (Q if R otherwise if S)". What does "otherwise if" mean? Seemingly it only means "or", so that the structure of the sentence would be "If P then (Q if (R or S))" which would be logically equivalent to "If P then (if (R or S) then Q)" which is equivalent to "If P and (R or S) then Q". But then plugging the predicates for P, Q, R and S back into the sentence, we have "If Opt(A,B) is an optional group pattern, and ( (S is a pattern solution of A and of B), or (S is a solution to A but not to A and B) ) then S is a solution of the optional graph pattern." And then this can be logically reduced to just "If Opt(A,B) is an optional group pattern, and S is a pattern solution of A, then S is a solution of the optional group pattern." But that does not make sense, because it seems that B has no role in defining the result. I think you mean the following: Let A and B be GraphPatterns. Let V be the set of variables and blank nodes in A OPTIONAL { B }. Let VA be the set of variables and blank nodes in A. Let VB be the set of variables and blank nodes in B that are not in VA. Let S be a partial map V to RDF-T, such that the following two conditions hold: a) for all v in VA, S(v) is defined (ie, S restricted to VA is a total map) b) for all v1, v2 in VB, if S(v1) is defined, then S(v2) is defined. That is, S restricted to VB is either a total map, or an empty map (or, put another way, the domain of S intersected with VB is either the empty set or VB.) Then S is a solution for A OPTIONAL { B } if S if the following two conditions hold: a) S is a solution for A and, b) if the domain of S intersected with VB is VB, then S is a solution for B. Fred ZemkeReceived on Thursday, 12 January 2006 17:32:12 GMT

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