Forcing filters versus Boolean filters

If $G$ is a forcing filter, nonemptiness and upward closure put $S\in G$. For $A,B\in G$, directedness supplies a nonempty $C\in G$ with $C\subseteq A\cap B$. Thus $A\cap B$ is a condition, and upward closure puts it in $G$. This proves Boolean filter closure, while $\mathrm{\varnothing }\mathrm{\notin }G$ holds because $G\subseteq P$. Conversely a proper Boolean filter contains $S$, hence is nonempty, and its meet $A\cap B\in G$ is a nonempty common stronger condition. The two notions therefore agree here.

For $S=\{0,1\}$ the conditions are $a=\{0\}$, $b=\{1\}$ and $t=\{0,1\}$, with $a,b\subseteq t$. A filter must contain $t$. It cannot contain both $a,b$, since $a\cap b=\mathrm{\varnothing }$ and there is no common stronger condition. The three possibilities are consequently $\{t\}$, $\{a,t\}$ and $\{b,t\}$, and each satisfies F1. The last two are maximal; adding either singleton to $\{t\}$ gives one of them.

The maximal families decide each of the four complementary pairs in the Boolean algebra, as F2 also predicts. The empty family fails F1, although its upward and pairwise-directed conditions alone would be vacuous. If $S$ is a singleton there is only the condition $t$ and the filter $\{t\}$. If $S=\mathrm{\varnothing }$, $P$ is empty and is excluded by the definition of forcing preorder; there is also no proper Boolean filter on $\mathcal{P}(S)$. Thus the correspondence spends no choice and supplies no nonexistent meets on arbitrary preorders. QED.