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Subject: Minutes: XDI TC Telecon Thursday 1-2PM PT 2009-05-07

Following are the minutes of the unofficial telecon of the XDI TC at:

Date:  Thursday, 07 May 2009 USA
Time:  1:00PM - 2:00PM Pacific Time (20:00-21:00 UTC)


Markus Sabadello 
Giovanni Bartolomeo
Drummond Reed
John Bradley

Bill Barnhill



We continued to discuss the new proposed definitions for these metagraph
predicates. See the following list messages for background:


We began by discussing Drummond's last post (the last link above). Drummond
explained that the distinction between $has and $has$a was particularly
useful in XDI dictionary definitions because it allowed the dictionary
author to explicitly declare:

1) When a dictionary subject is related to another dictionary subject in a
way that creates a new dictionary subject ($has).

2) When a dictionary subject has a property ($has$a).

For example:


The first set of $has dictionary statements produces the new dictionary


By contrast, the $has$a statements simply indicate that it +age,
+serial+number, and +color are all valid properties of +car.

Giovanni asked if the same subject could the object of both a $has and a
$has$a statement from another subject. For example:


Drummond agreed it could. The difference between the $has and $has$a
statements is that the $has statement is an assertion that there exists
another XDI RDF subject that serves to identify the new set, whereas the
+x/$has$a/+y statement does NOT assert that there exists another XDI RDF
subject that identifies the new set, is only asserts that the property +y is
a valid property on +x.

Giovanni felt that for any property +y on a subject +x, there may be the
need to talk about the set of objects that have that property on that
subject, and that implies +x+y (+x/$has/+y).

Drummond agreed that this implication can always be made, and thus an XDI
context that had an instance of +x/+y/... could always be queried about
+x+y. However that doesn't solve the use case where you just want to query
+x to find out: a) what subsets it has, and b) what properties it has. $has
and $has$a satisfy these use cases because you can do the query:


It was agreed to continue discussion on the mailing list.

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