I have to construct a proof related to the notion of filter. This is the problem:
"$F$ is a filter on a Boolean algebra $B$. Show that $F$ is a prime filter on $B$ if and only if for all $x\in B$, $x\in F$ or $\lnot x\in F$."
I know that the filter $F$ is a prime filter if
$F$ is not equal to $B$
if $s\lor t\in F$, then $s\in F$ or $t\in F$.
Since I have a "if and only if" statement, I'll have to prove something "two ways", but I am not sure how to start. It's a Boolean algebra, so maybe I could use the fact that $a \lor \lnot a = 1$ and $a\land \lnot a = 0$?
The property should be more accurately stated as
$$\forall x \in B: x \in F \text{ or } \lnot x \in F \textit{ but not both }\tag{1}$$
$1 \in F$ for any filter $F$ and $1 = x \lor \lnot x$ so a prime filter satisfies $(1)$, as it cannot contain both (because then $0 = x \land \lnot x \in F$ and then $F=B$ which cannot be.)
If $F$ satisfies $(1)$ and we have $s,t \in B$ with $s \lor t \in B$. If $s \notin F$ and $t \notin F$ (so we're striving for a contradiction), the property tells us that $\lnot s \in F$ and $\lnot t \in F$, but then $\lnot s \land \lnot t =\lnot(s \lor t) \in F$, and so $s \lor t$ and its complement both are in $F$ and this cannot happen by $(1)$. This contradiction concludes the proof.