This post not CC'd by email On Wed, 01 Dec 1999 16:39:02 -0400, Gus Gassmann <Horand.Gassmann@dal.ca> wrote:
>1. Since (L,200) = 1 and (G,200) = 1, (L,G) must be 1. > >2. Since (L,G) = 1 and (L,F) = 0, (F,G) = F(200) = 0. > >3. Therefore (C,F) = 1, and also (G,N) = 1.
It is fabulous to have a second desription. One visual like Karnaugh maps and the other Boolean algebra. It is all coming together nicely. I wonder what other surprises you solution oriented people can throw at it.