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Subject: RE: [xdi] A question on semantics

• From: "Barnhill, William [USA]" <barnhill_william@bah.com>
• To: "kari@bitworld.us" <kari@bitworld.us>, "xdi@lists.oasis-open.org" <xdi@lists.oasis-open.org>
• Date: Thu, 24 May 2012 23:24:03 +0000

```> On 2012-05-24 6:38 AM, Kari Lippert wrote:
>
> Why would you infer that =bob/\$is/=charlie ?
>

That's because it is transitive.
=alice+friend/\$is/=bob isn't the right statement to express that =bob is +friend of =alice, from the explanation Markus gave, so you wouldn't actually have that.

That would be =alice/+friend/=bob.

I have another question for Markus or others...
What statements would assert that =alice is a +person, and any +person may have one or more +friend arcs, each of which point to an XRI?

>
> On 2012-05-24 10:03 AM, Markus Sabadello wrote:
> > 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 [3]> 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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```

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