Re: [xdi] Proposal for separation of subject/predicate and parent-subject/child-subject semantics

[Drummond Reed <drummond.reed@xdi.org>]

Giovanni,

Two things:

1) First, I sent email yesterday modifying my proposal to use subsegment delimiters as predicates to instead propose using $has$a as the predicate for expressing parent-subject/child-subject relationships (+a+b), and dedicated $has for expressing subject/predicate relationships. So that means:

+a/$has/+b  <==>  +a/+b
+a/$has$a/+b  <==>   +a+b

2) The whole point of my proposal was to separate the two statements above so that +a/+b does NOT imply that +a+b exists in the same graph and vice versa. As Markus said in his message, they are different graphs. The fact that I have =drummond/+email$1 does not mean the same graph contains =drummond+email$1 and vice versa.

They are two different statements. In particular, if you look at the "golden triangle" of metagraph relationships shown below, saying $has$a (which is literally $has/$has/$a) defines a subset of $has relationships in which the predicate is also an object. Asserting a predicate as an object is different that asserting it as a predicate. Neither implies that the other exists. They do have a logical relationship - it is the same predicate involved in both cases (+email$1), but asserting it as a predicate does not mean it is also an object, and asserting is as an object does not need it is also a predicate.

=Drummond

On Tue, Apr 20, 2010 at 8:21 AM, Giovanni Bartolomeo <giovanni.bartolomeo@uniroma2.it> wrote:

Dear Drummond, All,

thank you for this proposal - two main thoughts from my side:

1) it is FUNDAMENTAL that if +a/+b is resolvable then also +a+b MUST be, and viceversa. We used the symbol <==> to say this, +a/+b <==> +a+b.
Obviously the stored document can only contain one of the two, e.g. the document only contains +a/+b, but +a+b MUST be resolvable as well.
This is about the relationship between asserted and inferred statements which I illustrated in the second part of my email http://lists.oasis-open.org/archives/xdi/201004/msg00004.html - we didn't discuss this yet because we ran out of time last week, but it would be good to address this in next phc.

2) About the following statements (in which the symbol <==> is used in a somehow different way than in point 1)

+a/$has/+b      <==>    +a/+b
+a/+/+b <==>    +a+b
+a/$/$b <==>    +a$b
+a/=/=b <==>    +a=b
+a/@/@b <==>    +a@b
+a/*/*b <==>    +a*b
+a/!/!b <==>    +a!b
+a/()/(b)       <==>    +a(b)

note that the statements on the left side in their own give origin to metastatements like +a/$has/+, +a/$has/$, +a/$has/() which in turns are logically equivalent to +a+, +a$, +a() whose semantics is totally unknown to me.

Kind Regards,
Giovanni

Def. Quota "Drummond Reed" <drummond.reed@xdi.org>:

Per my action item from Thursday's telecon, following is more about my
analysis and proposal for solving the "$has semantics problem" once
and for all.

*** ANALYSIS ***
My first conclusion, on which I think we all agree, is that while +a+b
might imply +a/+b and vice versa, that does not mean that: a) they
have the same semantics, or b) that they identify the same XDI graph.
Indeed, since one is an XDI subject and the other adds an XDI
predicate, by definition they cannot identify the same XDI graph.

To illustrate in English, most English verbs can be made into nouns by
putting them in their infinitive form (“to be”, “to run”, “to shout”).
But “run” and “to run”, while clearly linked semantically, do not have
the same meaning. They cannot be interchanged with each other.

Thus my second conclusion is that none of the three XDI graph examples
we were discussing on our telecon two weeks ago were in fact
equivalent. They were each different XDI graphs that had similar but
not identical semantics.

My third conclusion is that this means that we do in fact need clearly
distinguished ways of semantically and syntactically of expressing
that an XDI subject (e.g., +a) : 1) has a child subject (e.g., +a+b),
and 2) has a predicate (e.g., +a/+b).

My fourth conclusion was that this is the problem we have been
struggling with for the last year: by using $has (with or without
variants like $has$a) to express both parent subject/child subject
relationships (+a+b) and subject/predicate relationships (+a/+b), we
have wrapped ourself in a classic Gordian knot.

*** PROPOSAL ***

Once the problem was framed this way, it became much easier (for me
anyway) to see that a clean solution would be to dedicate the $has
predicate to only one semantic – either the parent subject/child
subject relationship or the subject/predicate relationship. Of those
two options, I strongly prefer dedicating the $has verb to expressing
the subject/predicate relationship (+a/+b), since this is the simpler
and more intuitive to understand, and mirrors the semantics of the $a
predicate, which expresses predicate/object relationships.

If $has is dedicated to that expression, then we simply need a
different predicate (or set of predicates) to express parent
subject/child subject relationships.

For that, an obvious answer is to use XRI delimiters as predicates.
This is intuitive since it is syntactically required to have a
delimiter between XRI subsegments, and each of the six XRI delimiters
(not counting parentheses for cross-references) has its own
XRI-defined "semantics", e.g., * means reassignable identifier, !
means persistent identifiers, etc.

So to summarize using +a and +b notation:

+a/$has/+b      <==>    +a/+b
+a/+/+b <==>    +a+b
+a/$/$b <==>    +a$b
+a/=/=b <==>    +a=b
+a/@/@b <==>    +a@b
+a/*/*b <==>    +a*b
+a/!/!b <==>    +a!b
+a/()/(b)       <==>    +a(b)

Semantically, it would still be true that +a+b/$is$a/+b (and the
inverse that +b/$a/+a+b). And while +a+b would imply that +a/+b is
valid, and vice versa, the existence of one in an XDI graph would not
imply the existence of the other.

So, if =example wanted to do a query to determine if +a+b and +a/+b
both existed in an XDI graph, it would be:
=example
       $get
               /
                       +a
                               $has
                                       +b
                               +
                                       +b

I have been trying out this proposal in sample XDI documents for the
past week and it seems to work very well. Please consider and
experiment with it and post your thoughts. We'll put it on the agenda
for next week's telecon.

=Drummond

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