Talk:De Morgan's Laws imply Uniquely Complemented Lattice is Boolean Lattice

Definitely worth reading on this topic: --Dfeuer (talk) 06:27, 22 January 2013 (UTC)

The relation $a \preceq b \iff a \wedge b = a$ is used and derived in e.g. Equivalence of Definitions of Lattice. You may want to use Ordering in terms of Meet (and its dual Ordering in terms of Join) which I haven't come round to yet but have referred to a few times. --Lord_Farin (talk) 08:35, 22 January 2013 (UTC)

For $(3)$ implies $(1)$:

Let $x \in S$. Then:

From Poset Elements Equal iff Equal Weak Upper Closure it follows that:


 * $\neg \left({p \wedge q}\right) = \neg p \vee \neg q$

as desired.

I hope this appeals to you more. --Lord_Farin (talk) 07:55, 23 January 2013 (UTC)


 * I'd like to have both your approach and an improved version of mine. One thing I've been pondering (because I'm so terribly ignorant) is the relation between Boolean lattices and Boolean algebra. It looks like if you take a set of logic variables (or whatever) you can build a Boolean lattice out of them. Then $\vee$ and $\wedge$ will have their usual logical meanings, $\top$ and $\bot$ will be true and false, and $\preceq$ will correspond to $\implies$. Your proof seems a bit weird to me, in that context, since the weak upper closure of a proposition is the set of all propositions in the lattice that that proposition implies. It brings in more of the structure of the lattice, I believe. --Dfeuer (talk) 20:30, 23 January 2013 (UTC)


 * Your intuition is right; that it works is the very aim of the definition of BL. If you have trouble relating to the definition, you can always substitute either of the to-be-equal expressions for $x$ and derive it that way - such obscures the generality of the argument, however. I think it's good to have both arguments. --Lord_Farin (talk) 20:34, 23 January 2013 (UTC)


 * I just like the fact that my style of argument seems (I believe) to limit itself to logical methods. It should be possible to formulate the exact same argument to prove something like $\neg (a \lor b) \dashv\vdash \neg a \land \neg b$ from $a \implies b \dashv\vdash \neg b \implies \neg a$ (if I'm using the notation even vaguely correctly). Your argument seems to reach to another level (universe of discourse?) in order to form the notion of upper closure. Or, then again, maybe I'm talking complete nonsense out of ignorance. --Dfeuer (talk) 21:16, 23 January 2013 (UTC)


 * That said, we should have both approaches. --Dfeuer (talk) 21:22, 23 January 2013 (UTC)