> > Indeed, the expressions should terminate. Since you can include a function call in your path expression, and you can include your own custom function, there is no guarantee that the expression will terminate.
I.e. you need to take a subset of possible path expression already, in general case the problem is undecidable. > > 2016-01-27 11:11 GMT+01:00 Pavel Velikhov <[email protected] > <mailto:[email protected]>>: > >> >> Or simplified: is the set selected by p1 equal to the set selected by p2? > > If we allow p1 and p2 to be arbitrary XQuery path expressions, then its > undecidable. > I can reduce the halting problem of Turing machines to this test. > >> >> 2016-01-27 11:09 GMT+01:00 W.S. Hager <[email protected] >> <mailto:[email protected]>>: >> Isn't the constraint in this case the test: is p2 a subset of p1? >> >> 2016-01-27 11:04 GMT+01:00 Pavel Velikhov <[email protected] >> <mailto:[email protected]>>: >> >> > On 27 Jan 2016, at 12:54, W.S. Hager <[email protected] >> > <mailto:[email protected]>> wrote: >> > >> > Can't we formally proof something as obvious Adam's case? >> >> In Adam’s case we want to test whether path expression p1 subsumes path >> expression p2. >> If we don’t put any conditions on p1 and p2, the problem is undecidable: p1 >> and p2 may include >> function calls, so the expressive power of p1 and p2 are that of a Turing >> Machine. >> >> >> >> -- >> W.S. Hager >> Lagua Web Solutions >> http://lagua.nl <http://lagua.nl/> >> >> >> >> -- >> W.S. Hager >> Lagua Web Solutions >> http://lagua.nl <http://lagua.nl/> > > > > > -- > W.S. Hager > Lagua Web Solutions > http://lagua.nl <http://lagua.nl/>
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