# Re: A different definition of MINUS, Part 3

Date: Mon, 22 Dec 2008 09:23:14 -0800 (PST)

>
>
>
> >>>>Projection, to me, doesn't seem like any sort of union.
>
> >>>OK, in classic relational algebra union can only be applied to the
> >>>relations with the same header (that is set of attributes). Therefore,
> >>>when generalizing union to become applicable to any pair of relations
> >>>one must decide first, what the header the resulting relation should
> >>>have. D&D assumed it has to also be a union, but I suggest that it can
> >>>be anything: intersection, difference, or even symmetric difference.
> >>>However, the last two choices are no good: symmetric difference would
> >>>make the generalized version of the union incompatible with classic RA
> >>>union, while difference operation is not symmetric, thus rendering
> >>>generalized union nonsymmetric as well. Therefore, the only
> >>>alternative to D&D version of the union is  "inner union": it
> >>>intersects over headers, and unions over tuples. Compare it to join
> >>>that intersects on tuple level, and unions headers.
>
> >>>Next one may compare D&D <AND>&<OR> based system, with RL join&inner
> >>>union based one in terms of consistency. Both have arguments in their
> >>>favor. D&D system honors distributivity, and De Morgan laws. RL honors
> >>>absorption, so that the subset relation can be generalized to be
> >>>applicable to any pair of relations. Also RL can express projection as
> >>>an (inner) union of a relation with an empty relation. First, tuples
> >>>in both relations (there are none in the second!) are collapsed to the
> >>>common set of attributes. These are essentially projections. Then we
> >>>make a union of projections, but keep in mind that the second
> >>>projection is empty!
>
> >>Okay, you seem to be saying that DeMorgan holds for D&D but not for RL.
> >>Didn't you use DeMorgan in the proof that amazed you? Was that D&D or RL?
>
> > In RL the <AND>  operator is not considered fundamental operator, and
> > can be represented
> > *algebraically* as
>
> > x <AND> y ≝ (x ^ (y v R11)) v (y ^ (x v R11)).
>
> > This operator can be proven enjoying some nice properties, including
> > DeMorgan.
>
> When you used DeMorgan, I didn't think you were using <AND>. Weren't you
> using ^ and + ?

This was a typo.
http://arxiv.org/ftp/arxiv/papers/0807/0807.3795.pdf page 5 Received on Mon Dec 22 2008 - 18:23:14 CET

Original text of this message