Do Boolean Functions connect set operations and composite sentences? I've been looking for a connection between composite sentences and set operations, is the link Boolean Functions? This is of course assuming the number of component statements is equal to the number of subsets of U.
 A: First we take any 'universe' $U$. Then indeed operations on sets contained in $U$ correspond directly to the boolean operations simply by definition, and in fact the subsets of $U$ under set inclusion form a boolean lattice.
$
\def\xor{\oplus}
\def\wi{\subseteq}
\def\less{\smallsetminus}
\def\diff{\mathbin{Δ}}
\def\power{\mathcal{P}}
\def\ii{\mathbf{1}}
$

Take any sets $A,B \wi U$.
$A \cup B = \{ x : x \in A \lor x \in B \}$.
$A \cap B = \{ x : x \in A \land x \in B \}$.
$U \less A = \{ x : x \in U \land \neg x \in A \}$.
$A \diff B = \{ x : x \in U \land ( x \in A \xor x \in B ) \}$
$A \wi B$ iff $\forall x \in U\ ( x \in A \to x \in B )$.
Take any $c \in U$.
$\ii_{A \cup B}(c) \equiv \ii_A(c) \lor \ii_B(c)$.
$\ii_{A \cap B}(c) \equiv \ii_A(c) \land \ii_B(c)$.
$\ii_{U \less A}(c) \equiv \neg \ii_A(c)$.
$\ii_{A \diff B}(c) \equiv \ii_A(c) \xor \ii_B(c)$.
$A \wi B$ iff $\ii_A(x) \to \ii_B(x)$ for every $x \in U$.

$\cup,\cap,\diff$ readily generalize to an arbitrary finite number of sets, because $\lor,\land,\xor$ are associative. Furthermore, $\cup,\cap,\less$ generalize to arbitrary families of sets as follows.

Take any $F \wi \power(U)$.
$\bigcup F = \{ x : x \in U \land \exists A \in F\ ( x \in A ) \}$.
$\bigcap F = \{ x : x \in U \land \forall A \in F\ ( x \in A ) \}$.
$U \less \bigcup_{A \in F} A = \{ x : x \in U \land \neg \exists A \in F\ ( x \in A ) \}$
$\ = \{ x : x \in U \land \forall A \in F\ ( \neg x \in A ) \} = \bigcap_{A \in F} ( U \less A )$.
Take any $c \in U$.
$\ii_{\bigcup_{A \in F} A}(c) = \bigvee_{A \in F} \ii_A(c)$.
$\ii_{\bigcap_{A \in F} A}(c) = \bigwedge_{A \in F} \neg \ii_A(c)$.
$\ii_{U \less \bigcup_{A \in F} A}(c) = \neg \bigvee_{A \in F} \ii_A = \bigwedge_{A \in F} \neg \ii_A(c) = \ii_{\bigcap_{A \in F} ( U \less A )}(c)$.
