Generated filters and the complementary-pair tests

The displayed set contains $1$ using $n=0$, contains each $s\in S$ using $n=1$, and is upward closed. Concatenating two finite lists witnessing membership of $b,c$ gives a finite meet below $b\wedge c$, so it is a filter. Any filter containing $S$ contains each finite meet, including $1$, and therefore every element of the displayed set. This proves minimality, including $⟨\mathrm{\varnothing }⟩=\{1\}$.

Suppose $U$ is maximal proper and $b\mathrm{\notin }U$. Then $⟨U\cup \{b\}⟩$ is improper. Step 1.1 gives $u\wedge b=0$ for some $u\in U$, combining all old factors into $u$, with $u=1$ if there are none. Consequently $u\le \mathrm{\neg }b$, so $\mathrm{\neg }b\in U$. A proper filter cannot contain both $b$ and $\mathrm{\neg }b$, since their meet is $0$. Conversely, if $U$ decides every complementary pair, any larger filter contains some $b\mathrm{\notin }U$ together with $\mathrm{\neg }b\in U$ and is improper.

A prime filter decides complementary pairs because $1=b\vee \mathrm{\neg }b\in U$. Conversely, if a proper filter decides pairs and neither $b$ nor $c$ belongs to it, both complements do, so $\mathrm{\neg }(b\vee c)=\mathrm{\neg }b\wedge \mathrm{\neg }c$ belongs to it; thus $b\vee c\mathrm{\notin }U$. This proves primality from pair decision.

Filter closure and upward closure give $b\wedge c\in U$ exactly when both $b,c\in U$; primality gives $b\vee c\in U$ exactly when at least one belongs to $U$. Pair decision gives ${\chi }_U(\mathrm{\neg }b)=1-{\chi }_U(b)$, and properness gives ${\chi }_U(0)=0$, ${\chi }_U(1)=1$. These are precisely all homomorphism identities into $2$. Conversely a bound-preserving homomorphism indicator satisfies the complementary-pair identity, giving pair decision.

The operation $F\mapsto \{\mathrm{\neg }f:f\in F\}$ is an inclusion-preserving bijection from filters to ideals: complementation reverses inequalities, sends meets to joins, and exchanges $1$ with $0$; applying it twice returns $F$. It preserves properness and maximality. For a pair-deciding $U$ its image is $B\setminus U$. This complement is prime because $a\wedge b\mathrm{\notin }U$ implies at least one of $a,b$ is not in $U$. Conversely, for a prime ideal $I$, primality applied to $b\wedge \mathrm{\neg }b=0$ and properness give exactly one of $b,\mathrm{\neg }b$ in $I$. Its complement is consequently a pair-deciding filter, hence an ultrafilter. Maximal ideals correspond to maximal filters by the bijection, completing both assertions. The trivial algebra has no proper filters or ideals, so those quantified assertions are vacuous there. QED.