$X\setminus (a\cup b)=(X\setminus a)\cap (X\setminus b)$ and $X\setminus (a\cap b)=(X\setminus a)\cup (X\setminus b)$; and for a nonempty set $F$, $X\setminus \bigcup F=\bigcap \{X\setminus a:a\in F\}$ and $X\setminus \bigcap F=\bigcup \{X\setminus a:a\in F\}$

Every $X\setminus a$ is a subset of $X$, so the class $\{X\setminus a:a\in F\}$ is the separated set $\{w\in \mathcal{P}(X):\mathrm{\exists }a(a\in F\wedge w=X\setminus a)\}$, hence a set; and it has a member because $F$ does.

$z\in X\setminus (a\cup b)$ holds exactly when $z\in X$ and neither $z\in a$ nor $z\in b$, that is, exactly when $z\in X$ and $z\mathrm{\notin }a$, and $z\in X$ and $z\mathrm{\notin }b$; this is $z\in (X\setminus a)\cap (X\setminus b)$.

$z\in X\setminus (a\cap b)$ holds exactly when $z\in X$ and not both $z\in a$ and $z\in b$, that is, exactly when $z\in X$ and $z\mathrm{\notin }a$, or $z\in X$ and $z\mathrm{\notin }b$; this is $z\in (X\setminus a)\cup (X\setminus b)$.

$z\in X\setminus \bigcup F$ holds exactly when $z\in X$ and no member $s$ of $F$ has $z\in s$, that is, exactly when $z\in X\setminus a$ for every $a\in F$; the sets $X\setminus a$ with $a\in F$ are precisely the elements of $\{X\setminus a:a\in F\}$, which is nonempty, so this is $z\in \bigcap \{X\setminus a:a\in F\}$.

$z\in X\setminus \bigcap F$ holds exactly when $z\in X$ and $z$ fails to lie in some member $a$ of $F$, that is, exactly when $z\in X\setminus a$ for some $a\in F$; this is $z\in \bigcup \{X\setminus a:a\in F\}$.

The two finite laws and the two family laws are established, which is the statement.