[xquery-talk] Function for determining one XPath as subset of another

[W.S. Hager]

Well, so, to continue, let's assume that there are no user-defined
functions, and in fact the only thing we want to proof is select+filter,
where a filter is limited to the default operators. From that is it follows
that

-path1:
select-child-nodes-by-name(select-child-nodes-by-name($context,'x'),'w')
-path2: select-descendant-nodes-by-name($context,'w')

And we need to proof that those functions are somehow equivalent. Can it be
done?

Op woensdag 27 januari 2016 heeft daniela florescu <dflorescu at me.com> het
volgende geschreven:

> >
> > It seems to be a long-standing tradition that computer scientists, when
> asked to prove a difficult conjecture C, respond by giving a proof for a
> simplified conjecture C'. While this might lead to progress in the long
> run, and enables them to get papers published in the academic literature,
> it is totally useless to practical engineeers who want to know whether they
> can safely rely on C.
>
> Michael,
>
>
> what you say is nice and true.
>
> However given that:
> 1. path expressions point (syntactically hence esemantically) into
> XQuery’s expressions
> 2. XQuery expression language is Turing complete
> 3. Subsumption for a Turing complete language is undecidable.
>
> Well, I can hardly see a way to decide this problem other then by
> introducing SOME restrictions
> of some sort… but of course some restrictions that would not nullify the
> original problem all together
> and make the solution useless.
>
> Best,
> Dana



-- 

W.S. Hager
Lagua Web Solutions
http://lagua.nl
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