Re: Distributivity in Tropashko's Lattice Algebra
Date: 17 Aug 2005 09:32:24 -0700
Message-ID: <1124296344.878256.250480_at_o13g2000cwo.googlegroups.com>
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)
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 - 18:32:24 CEST