> Federico, I think the documentation in node-position.xq needs some more work.
> I can do this, but first we must agree on what is exactly the functionality
> and the assumptions about node positions that you rely on. For example, in the
> documentation of the np:node-position() function, you write:
> "The returned URI is stable, i.e. it does not change when other nodes are
> inserted, deleted or modified"
> Do you rely on this? If not, then we shouldn't write it down.

Yes, at the moment I am relying on it. It could be possible however, with some 
limitations, to lift off this requirements. I have not a clear idea at the 
of the limitations involved or the amount of work required to do so. If you 
is bad to promise that uri are stable I can look into it.

> Also, look at the following rewrite of the ancestor-of function. Do you agree
> with it?
> (:~
>  : Determines whether the node position that is given as second argument is
>  : an ancestor of the node position is given as first argument.
>  :
>  : If the two positions were obtained within the same snapshot S, then the
>  : result of the function applies to the corresponding nodes as well, that
>  : is, within snapshot S, the second node is an ancestor of the first.
>  : Otherwise, the result of the function does not imply anything about the
>  : positional relationship of the two nodes.
>  :
>  : @param $pos1 the potential descendant node position
>  : @param $pos2 the potential ancestor node position
>  :
>  : @return true if the node position $pos2 is an ancestor of the node position
>  : $pos1; false otherwise.
>  :
>  : @error zerr:ZAPI0028 if one of the given URI is not a valid node
>  : position computed by the <tt>np:node-position</tt> function.
>  :)
> declare function np:ancestor-of(
>   $pos1 as xs:anyURI,
>   $pos2 as xs:anyURI) as xs:boolean external;

Yes, I agree with it. When I wrote this I was a little scared to say that a 
structural relationship was ancestor of another, but now that they are called
node positions it should be ok. I am doing it.


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