User:Dfeuer/Implication in Boolean Lattices

Theorem

Let $(L, \wedge, \vee, \preceq)$ be a Boolean lattice.

Let $a, b \in L$.

Then $a \preceq b \iff \neg a \vee b = \top$.

Proof

Suppose $a \preceq b$.

Then $a \wedge b = a$.

Thus $\neg a \vee \neg b = \neg a$.

So $\neg a \vee b = (\neg a \vee \neg b) \vee b = \top$.

Suppose $\neg a \vee b = \top$.

Then $a \wedge (\neg a \vee b) = (a \wedge \neg a) \vee (a \wedge b) = a \wedge b = a$.

Thus $a \preceq b$.

$\blacksquare$