Re: Distributivity in Tropashko's Lattice Algebra

Marshall Spight wrote:
> Mikito Harakiri wrote:
> >
> > (a join c)
> > union
> > (b join c)
> > union
> > (a join b) = b union c
> >
> > This single equation is equivalent to both predecessors. Indeed to
> > derive the first one, join both sides with "c". On left side leverage
> > distributivity, and on the right side apply absorption law. This gives
> >
> > (a join c)
> > union
> > (b join c)
> > union
> > (a join b join c) = c
>
>
> Mikito,
>
> I have reread your post a few times. I follow all of it except the
> above. I don't see how you can apply distributivity and absorbtion
> at the same time and go from the first to the second. (I think I
> am only capable of understanding proofs that go one step at a time.
> If that.)

Gluing several steps into one is certainly a sloppy way to cause confusion:-)

1. Join both sides with "c":

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

2) Absorption law on the right:

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

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

((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 - 19:04:12 CEST