Talk:Monadic Boolean algebra
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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. ∃(x + y) = ∃x + ∃y;
- 4. ∃x∃y = ∃(x∃y)
- vs.
- 3. ∀(xy) = ∀x∀y;
- 4. ∀x + ∀y = ∀(x + ∀y)
- Okay, we've got
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)
- 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)
-
- Okay, I get it. Vivacissamamente 17:27, 28 October 2005 (UTC)
