Re: First order logic and SPARQL

Date: Sun, 5 Sep 2010 14:29:36 -0500
Message-ID: <AANLkTinaA5v8jAFOJkmj3Jzw12=pzJe48p61Ggg-3cQZ@mail.gmail.com>
To: Bob MacGregor <bob.macgregor@gmail.com>

```Bob --

You may be interested in a version of* recursive datalog +
negation-as-failure* that has a (unique) model-theoretic semantics.

It's described in [1], and forms the basis for reasoning in [2].

[1]  Backchain Iteration: Towards a Practical Inference Method that is
Simple
Enough to be Proved Terminating, Sound and Complete. Journal of Automated
Reasoning, 11:1-22

A Wiki and SOA Endpoint for Executable Open Vocabulary English Q/A over SQL
and RDF
Online at www.reengineeringllc.com

Reengineering

On Sat, Sep 4, 2010 at 11:10 AM, Bob MacGregor <bob.macgregor@gmail.com>wrote:

> I find this statement potentially misleading:
>
>
>         "SPARQL and non-recursive safe Datalog with negation have
> equivalent expressive power, and hence,
>          by classical results, SPARQL is equivalent from an expressiveness
> point of view to Relational Algebra"
>
> SPARQL, to its detriment, does not have a model-theoretic semantics
> (whereas logic languages like CommonLogic
> do).  The most obvious difference is that in logic, the AND operator is
> commutative, while in SPARQL, the
> order of conjuncts in an AND (a ".") makes a difference -- commute them,
> and you sometimes change the
> result/meaning of the query.
>
> My impression is that Datalog is in fact declarative (unlike Prolog).  I
> suppose its possible that a declarative language,
> nrs Datalog wn, and a non-declarative one, SPARQL, could have the same
> expressive power, even though
> they cannot be equated semantically (on the surface, that seems
> counterintuitive).  On the other hand, I'm
> wondering if you have somehow "dressed up" SPARQL to make it more
> principled than it really is to make your
> claim of "equivalence" -- are you talking about the real SPARQL, or some
> idealized version?
>
> - Bob
>
```
Received on Sunday, 5 September 2010 19:30:05 UTC

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