Re: <OR> predicate?

On Sep 27, 10:57 am, paul c <toledobythe..._at_oohay.ac> wrote:
> On 26/09/2010 4:11 PM, paul c wrote:
> ...
>
>
>
> > ps:There might be occasional usefulness in making what one might call
> > 'domain assertions', eg., in D&D Algebra, "there is a position called
> > 'toilet scrubber'" could be assessed from R{position} <OR> (<NOT>
> > R{position})...
>
> That makes me think of a question that may not have any practical point
> except for understanding the 'A-algebra' (because it involves
> non-union-compatible relations).
>
> Suppose the predicate of R{Position} is "position Position is occupied".
>
> I would think one possible predicate of R <OR> (<NOT> R) would be
> something like "position Position is occupied OR unoccupied".
>
> Seems that an even simpler expression, R <OR> TABLE_DEE, gives the same
> extension. Is the predicate the same? Or is there a good reason to
> think instead of the predicate as something like "position Position Exists"?
>
> (I'm asking this question even though I personally have some difficulty
> reconciling parts of the A-algebra formal definitions, eg., on one hand,
> the heading of R <OR> TABLE_DEE must include the heading of TABLE_DEE
> which is the empty set (in other words, the empty set is a
> member/element of the heading and I presume being a member is not the
> same as being a subset). On the other hand, the definition of an
> A-algebra heading says that it is a set of ordered pairs. But the empty
> set is certainly not an ordered pair. I assume I must be making some
> mistake, otherwise R <OR> TABLE_DEE is not a valid expression. By
> 'valid' I mean theoretically possible. Maybe somebody can point out how
> I'm making this mistake.)

This assertion (and any other) can be checked in QBQL; here is how:

1. Translate Tutorial D terms into QBQL: TABLE_DEE = R01 <OR> = _at_v (former "+", see http://vadimtropashko.wordpress.com/relational-programming-with-qbql/udf/) <NOT> = _at_' (former "'")
2. Write your assertion in QBQL x _at_v (x @') = x @v R01.
3. Run the program containing this assertion.

QBQL will iterate through all the relations in the database trying to find counterexample. In this case it finds none. Received on Mon Sep 27 2010 - 20:22:05 CEST