Re: [xdi] Realizations about $has statements

Hello Markus and Drummond,

I would propose the following:

=example
             $is$a
                         +person
=example+email
              $1
                         ??

=example+email is the set of emails belonging to =example, $1 is the  
accessor for ??. It is easy to find =example+email$1  
solving one by one the subsegments from left to right. And this is  
unambiguous, assumed that $has is left associative. What do you think?

Kind Regards,
Giovanni

Def. Quota "Markus Sabadello" <>:

> So the point is that in a PDS you want to do this, right?
>
> =example
>            $is$a
>                        +person
>            +email$1
>                        ??
> =example$1
>            $is$a
>                        +home+person
>            +home+email
>                        =example+email$1
>
> Not really sure how well it would work in practice.. If there were more
> subsegments than just =example and +email$1, how would I know where to break
> them up into their "subject" and "predicate" parts in order to follow the
> reference and find the literal value?
>
> I.e. in the above example, how do I know I have to look for subject =example
> and then for predicate +email$1, rather than for subject =example+email and
> then for predicate $1 ?
>
> Markus
>


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