There's the notion of co-inductive types. I think in Haskell you could proof the case, so why not in xquery?
2016-01-27 11:24 GMT+01:00 Pavel Velikhov <[email protected]>: > > > 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]>: > >> >> >> 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]>: >> >>> 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]>: >>> >>>> >>>> > On 27 Jan 2016, at 12:54, W.S. Hager <[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 >>> >> >> >> >> -- >> >> W.S. Hager >> Lagua Web Solutions >> http://lagua.nl >> >> >> > > > -- > > W.S. Hager > Lagua Web Solutions > http://lagua.nl > > > -- W.S. Hager Lagua Web Solutions http://lagua.nl
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