Re: Distributivity in Tropashko's Lattice Algebra

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.)

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.

a < b means a has a *not necessarily proper* subset of the attributes of b.
(This is a partial order, yes?)

It appears I originally managed to misinterpret

1) your terminology
2) your general approach
3) several individual statements

(Oops!)

Marshall Received on Wed Aug 17 2005 - 11:32:24 CDT