Consider a complete Boolean algebra $([A \to B], \leq_{[A \to B]})$ whose carrier set is the class of functions from the set $A$ to the set $B = \{1, 0 \}$, where $1$ and $0$ represent truth and falsity, respectively, where $(A, \leq_A)$ is a complete Boolean algebra.
Let $g \in [A \to B]$ be a function from $A$ to $B$. Suppose I want to define the smallest function $f \in [A \to B]$, such that, for all $x$, if $g(x) = 1$, then $f(x) = 1$ and $\forall S(S \subseteq \{x \mid g(x) = 1 \} \rightarrow f(\sqcup_{A} S) = 1$, where $\sqcup_A$ is the supremum operator of $(A, \leq_A)$. Can I express "the smallest function..." by using the supremum $\sqcup_{[A \to B]}$, as
$$\sqcup_{[A \to B]} \Bigl\{f \in [A \to B] \mid \text{for all }x\text{, if }g(x) = 1\text{, then }f(x) = 1\text{ and }\forall S(S \subseteq \{x \mid g(x) = 1 \} \rightarrow f(\sqcup_{A} S) = 1) \Bigr\} $$