Re: Distributivity in Tropashko's Lattice Algebra

One (obvious) step was still missing:-)

Mikito Harakiri wrote:
> 3) The lattice algebra of the header relations is distributive. Apply
> distributive law on the left:

Omitted brackets:

> (((a join c) join c)
> union
> ((b join c) join c)
> union
> ((a join b) join c)) = c
>

3a) Associativity:

((a join (c join c))
union
(b join (c join c))
union
((a join b) join c)) = c

> 4) Idempotence:
>
> ((a join c)
> union
> (b join c)
> union
> ((a join b) join c)) = c
>
> > Checking to make sure I understand your terminology:
> >
> > a, b, c are empty relations with the same headers (same set of
> > attributes)
> > as A, B, and C, respectively.
>
> Right.
>
> > a < b means a has a *not necessarily proper* subset of the attributes
> > of b.
>
> Well, formally
>
> a < b
>
> means
>
> a join b = b
> a union b = a
>
> For the sublattice of header relations (which is algebra of sets) it is
> indeed the same as the subset of attributes.
>
> > (This is a partial order, yes?)
>
> Yep
Received on Wed Aug 17 2005 - 19:13:05 CEST