RE: [xdi] A question on semantics

Ok, thanks Markus.

Here is slightly different one. What is the difference between (1) and (2):

(1) =alice +friend =bob

(2a) =alice () +friend
(2b) =alice+friend () =bob

Also, given what you said about $is, I am wondering what current convention is for denoting antonym (“is not”).

I seem to recall a push to use $is$a for that, which I was always against because it did not seem to make sense from an English perspective.

Is is current convention to use $is$a, $is$not, or something else entirely?

Kind regards,

Bill Barnhill

Booz Allen Hamilton - Belcamp,MD

1-443-924-0824| barnhill_william@bah.com

From: xdi@lists.oasis-open.org [mailto:xdi@lists.oasis-open.org] On Behalf Of Markus Sabadello
Sent: Thursday, May 24, 2012 10:04 AM
To: Barnhill, William [USA]
Cc: Michael Schwartz; Joseph Boyle; Yuriy Zabrovarnyy; OASIS - XDI TC; Cameron Hunt
Subject: Re: [xdi] A question on semantics

I think only (1) is correct.

(2) would mean that the _expression_ "Alice's Friend" is synonymous with "Bob", or something like that.

Also, according to my understanding, $is is commutative and transitive, i.e. if you had this:
=alice+friend/$is/=bob
=alice+friend/$is/=charlie

Then you could infer =bob/$is/=charlie, which I think is not the intention..

Markus

On Thu, May 24, 2012 at 3:09 PM, Barnhill, William [USA] <barnhill_william@bah.com> wrote:

Lastly, a question.
What is the semantic difference, if any, between the two statement sets below?

(1) =alice +friend =bob

(2a) =alice () +friend
(2b) =alice+friend $is =bob

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