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Subject: Inference and equivalence defintion/notation (and \$has)

• From: Drummond Reed <drummond.reed@xdi.org>
• To: OASIS - XDI TC <xdi@lists.oasis-open.org>
• Date: Fri, 15 Jan 2010 02:12:23 -0800

Bill,

In thinking about your definitions of "logical equivalence" and "resolution equivalence" today, I'm wondering if we couldn't just use the terms "inference" and "equivalence". In other words, it is easier for me to understand the differences between the sentence "XDI statement A infers XDI statement B" and the sentence "XDI statement A is equivalent to XDI statement B".

My understanding of inference is that it can be unidirectional (A infers B but B does not infer A) or bi-directional (A infers B and B infers A).

My understanding of equivalence is that it must be bidirectional, i.e., if A is equivalent to B then B MUST be equivalent to A. This is the purpose of XDI \$is statements – they express equivalence between an XDI subject and and XDI object.

If you agree these terms will work, let's use your proposed notation:

Unidirectional inference               <=        or         =>

Bidirectional inference                <=>

Equivalence                               =

Non-equivalence             /=

With that said I can restate the first part of the final conclusion of today's telecon (see http://lists.oasis-open.org/archives/xdi/201001/msg00046.html for the chat transcript) as saying that:

+a/\$has/+b  <=>  (+a/+b)  <=>  +a/+b  <=>  +a+b

This is true because each infers the other, and all these inferences are bidirectional (I think).

This would also imply that:

+a  <=>  (+a)

Again, this is because each implies the other (I have one small reservation about this but I'll save that for another time).

And part two of our final conclusion of the telecon was that:

+a  /=  (+a)

This is true because as you described resolving (doing a \$get) on +a and (+a) respectively would not return the same graph.

Lastly, we can finally get to the definition of \$has and \$has\$a statements. What I have been saying (but not expressing with the proper notation) is:

+a/\$has/+b  =  (+a/+b)  =  +a+b

+a/\$has/+b  /=  +a/+b

+a/+b  /=  +a+b

+a/\$has\$a/+b  =  +a/+b

+a/\$has\$a/+b  /=  (+a/+b)

Do you agree?

=Drummond

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