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

Michael Kay mike at saxonica.com
Wed Jan 27 02:50:39 PST 2016

To take an example, if someone proves that a particular equivalence holds on a simplified data model with no attribute, namespace, comment, or processing instruction nodes, then that is of absolutely no use to someone writing an optimizer for a product in which such nodes can be encountered.

Michael Kay

> On 27 Jan 2016, at 09:51, Pavel Velikhov <pavel.velikhov at gmail.com> wrote:
>> On 27 Jan 2016, at 12:37, Christian Grün <christian.gruen at gmail.com> wrote:
>>> Its a common practice for everybody, who needs to come up with formal proofs. You start with the most simplified definitions possible, that capture the essence of the problem.
>>> Then you get the skeleton of the proof that is hopefully very simple. Then you can add details back, hoping that the proof remains simple and tractable.
>>> So imagine starting the proof while considering all the possible variations of path expressions, all the Infoset stuff, all XML Schema details. I think its hopeless.
>> I’m completely in line with Michael’s observations. It would obviously
>> be nice to have proofs for the full rule sets of the discussed
>> languages; but as experience shows, no one will do it (the rare
>> exception might prove the rule), so we are stuck with the work that is
>> of limited practical use.
> I can’t agree with you. There are a lot of results that automatically hold for the full specification, even though they are
> proved on a clean and easy-to-use subset: undecidability and np-completeness or np-hardness for instance.
> For the full specifications its sometimes hard to grasp the semantics, so proving anything serious is impossible. 
> Example: try proving that SQL-2003 queries are equivalent to some superset of relational algebra.

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