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

[W.S. Hager]

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 <pavel.velikhov at gmail.com>:

>
>
> 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 <pavel.velikhov at gmail.com>:
>
>>
>>
>> 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 <wshager at gmail.com>:
>>
>>> 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 <pavel.velikhov at gmail.com>:
>>>
>>>>
>>>> > On 27 Jan 2016, at 12:54, W.S. Hager <wshager at gmail.com> 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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