Talk:Monadic Boolean algebra

dual[edit]

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

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

Okay, we've got

3. ∃(x + y) = ∃x + ∃y;

4. ∃x∃y = ∃(x∃y)

vs.

3. ∀(xy) = ∀x∀y;

4. ∀x + ∀y = ∀(x + ∀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 ∃3 we don't have any nested existentials, whereas in ∀4 we do; similarly, in ∀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)[reply]

Under duality, joins become meets besides ∃ becoming ∀ so for example the dual of x + y is xy and the dual of ∃x + ∃y is ∀x∀y etc Kuratowski's Ghost 15:40, 28 October 2005 (UTC)[reply]

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