Power Set with Union and Intersection forms Boolean Algebra

Theorem
Let $S$ be a set, and let $\mathcal P \left({S}\right)$ be its power set.

Denote with $\cup$, $\cap$ and $\complement$ the operations of union, intersection and complement on $\mathcal P \left({S}\right)$, respectively.

Then $\left({\mathcal P \left({S}\right), \cup, \cap, \complement}\right)$ is a Boolean algebra.

Proof
Taking the criteria for definition 1 of a Boolean algebra in turn:

$(BA \ 0):$ Closure
$\mathcal P \left({S}\right)$ is closed under both $\cup$ and $\cap$:
 * Power Set Closed under Intersection
 * Power Set Closed under Union
 * Power Set Closed under Complement

$(BA \ 1):$ Commutativity
Both $\cup$ and $\cap$ are commutative from Intersection is Commutative and Union is Commutative.

$(BA \ 2):$ Distributivity
Both $\cup$ and $\cap$ distribute over the other, from Union Distributes over Intersection and Intersection Distributes over Union.

$(BA \ 3):$ Identity Elements
Both $\cup$ and $\cap$ have identities:

From Power Set with Intersection is Monoid‎, $S$ is the identity for $\cap$.

From Power Set with Union is Monoid, $\varnothing$ is the identity for $\cup$.

$(BA \ 4):$ Complements
From Union with Complement:


 * $\forall A \in S: A \cup \complement \left({A}\right) = S$

which is the identity for $\cap$.

From Intersection with Complement:


 * $\forall A \in S: A \cap \complement \left({A}\right) = \varnothing$

which is the identity for $\cup$.

All the criteria for a Boolean algebra are therefore fulfilled.