Talk:Monadic Boolean algebra

dual
The two rules 3 aren't dual in this edition. How should they look? Vivacissamamente 12:29, 25 October 2005 (UTC)


 * Looks ok to me ??? Kuratowski's Ghost 22:18, 25 October 2005 (UTC)


 * Okay, we've got
 * 3. &exist;(x + y) = &exist;x + &exist;y;
 * 4. &exist;x&exist;y = &exist;(x&exist;y)
 * vs.
 * 3. &forall;(xy) = &forall;x&forall;y;
 * 4. &forall;x + &forall;y = &forall;(x + &forall;y)

Besides the fact that 3 and 4 seem to switch places, we also seem to have a difference of opinions in the nestings: namely, in &exist;3 we don't have any nested existentials, whereas in &forall;4 we do; similarly, in &forall;3 we don't have any nested universals, wheras in &exists;3 we do. I'm not sure which is right, but these don't look dual to me... if they are, please explain why. Thank you. Vivacissamamente 01:03, 28 October 2005 (UTC)


 * Under duality, joins become meets besides &exist; becoming &forall; so for example the dual of x + y is xy and the dual of &exist;x + &exist;y is &forall;x&forall;y etc Kuratowski's Ghost 15:40, 28 October 2005 (UTC)


 * Okay, I get it. Vivacissamamente 17:27, 28 October 2005 (UTC)