> "empty($A intersect $B//*)"
>
> XPath 1.0:
>
> "count($B/**) = count($B//* | $A)"
>
> Cheers, Wendell
>

Getting all the ancestors of $A is likely to be a lot faster than getting all
the descendants of $B, simply because a node typically has many descendants
but few ancestors.

Michael Kay
Saxonica