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

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

Every $a\cup b$ with $b\in F$ is a subset of $a\cup \bigcup F$, since an element of $b$ lies in $\bigcup F$; so $\{a\cup b:b\in F\}$ is the separated set $\{w\in \mathcal{P}(a\cup \bigcup F):\mathrm{\exists }b(b\in F\wedge w=a\cup b)\}$, hence a set, and it has a member because $F$ does.

$z\in a\cap (b\cup c)$ holds exactly when $z\in a$, and $z\in b$ or $z\in c$; that is exactly when $z\in a$ and $z\in b$, or $z\in a$ and $z\in c$, which is $z\in (a\cap b)\cup (a\cap c)$.

$z\in a\cup (b\cap c)$ holds exactly when $z\in a$, or $z\in b$ and $z\in c$; that is exactly when $z\in a$ or $z\in b$, and $z\in a$ or $z\in c$, which is $z\in (a\cup b)\cap (a\cup c)$.

$z\in a\cap \bigcup F$ holds exactly when $z\in a$ and $z\in s$ for some $s\in F$, that is, exactly when $z\in a\cap s$ for some $s\in F$; the sets $a\cap s$ with $s\in F$ are precisely the elements of $\{a\cap b:b\in F\}$, so this holds exactly when $z$ lies in some element of that set, which is $z\in \bigcup \{a\cap b:b\in F\}$.

$z\in \bigcap \{a\cup b:b\in F\}$ holds exactly when, for every $b\in F$, $z\in a$ or $z\in b$. If $z\in a\cup \bigcap F$ then either $z\in a$, and the condition holds outright, or $z$ lies in every element of $F$, and it holds again. Conversely, if the condition holds and $z\mathrm{\notin }a$, then $z\in b$ for every $b\in F$, so $z\in \bigcap F$; hence $z\in a\cup \bigcap F$ in both cases.

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