This is a basic question I'm trying to figure out: why the Boolean's filter definition corresponds to the order-theoretic definition of filter ?
Here follows the relevant definitions.
Definition 1 (Filter in a partially ordered-set)
A nonempty subset $F$ of a partially ordered-set $(P, \le)$ is a filter on $P$ if the
following conditions hold:
Definition 2 (Filter in a Boolean algebra)
A nonempty subset $F$ of a Boolean algebra $(\mathbb{B}, \vee, \wedge, \bot, \top, \neg)$ is a
filter if it is
Boolean algebra seen as a lattice can be presented with a relation order $\le$ with the following equivalence:
$x \le y \iff x \vee y = y \iff x \wedge y = x.$
My question is, under a Boolean algebra, why the two definitions of filter aforementioned are equivalent ?
It's pretty clear to me that the property of upward closure is equivalent as $x \vee y = \sup\{x, y\}$. But I have doubt about the first property (downward directed), especially the direction (Definition 1 $\implies$ Definition 2). For the reverse direction, we can take $z = x \wedge y$.
Draft:
$z \le x \textrm{ and } z \le y \iff z = z \wedge x \textrm{ and } z = z \wedge y$, and by substitution we have $z = z \land (x \land y)$. How can I prove that $z = x \land y$ and thus by hypothesis $x \land y \in F$ ?
