Theorem: Algebra of Sets is Boolean Algebra

An algebra of sets is a Boolean algebra.

Proof:
Let $\RR \subseteq \powerset S$ be a set $S$ upon which an algebra of sets has been constructed.
We identify:
{{begin-axiom}}
{{axiom | lc= Set union:
        | m = \cup
        | rc= with join $\vee$
}}
{{axiom | lc= Set intersection:
        | m = \cap
        | rc= with meet $\vee$
}}
{{axiom | lc= Relative complement:
        | m = \relcomp S a
        | rc= with complementation $\neg a$
}}
{{end-axiom}}
We demonstrate that the Boolean algebra axioms (formulation $2$) are fulfilled.
;{{Boolean-algebra-axiom|2|0|nolink}}:
From {{Algebra-of-sets-axiom|2}}:
:$\forall a, b \in \RR: a \cup b \in \RR$
From {{Algebra-of-sets-axiom|3}}:
:$\forall a, b \in \RR: \relcomp S a \in \RR$
Then:
{{begin-eqn}}
{{eqn | l = \relcomp S {a \cap b}
      | r = \relcomp S a \cup \relcomp S b
      | c = De Morgan's Laws: Complement of Intersection
}}
{{eqn | ll= \leadsto
      | l = \relcomp S {\relcomp S {a \cap b} }
      | r = \relcomp S {\relcomp S a \cup \relcomp S b}
      | c = taking relative complement of both sides
}}
{{eqn | ll= \leadsto
      | l = a \cap b
      | r = \relcomp S {\relcomp S a \cup \relcomp S b}
      | c = Relative Complement of Relative Complement
}}
{{eqn | ll= \leadsto
      | l = a \cap b
      | o = \in
      | r = \RR
      | c = {{Algebra-of-sets-axiom|2}} and {{Algebra-of-sets-axiom|3}}
}}
{{end-eqn}}
Hence:
:$\forall a, b \in \RR: a \cap b \in \RR$
and $\RR$ fulfils {{Boolean-algebra-axiom|2|0}}.
{{qed|lemma}}
;{{Boolean-algebra-axiom|2|1|nolink}}:
We have:
{{begin-eqn}}
{{eqn | q = \forall a, b \in \RR
      | l = a \cup b
      | r = b \cup a
      | c = Union is Commutative
}}
{{eqn | l = a \cap b
      | r = b \cap a
      | c = Intersection is Commutative
}}
{{end-eqn}}
and $\RR$ fulfils {{Boolean-algebra-axiom|2|1}}.
{{qed|lemma}}
;{{Boolean-algebra-axiom|2|2|nolink}}:
We have:
{{begin-eqn}}
{{eqn | q = \forall a, b, c \in \RR
      | l = a \cup \paren {b \cup c}
      | r = \paren {a \cup b} \cup c
      | c = Union is Associative
}}
{{eqn | l = a \cap \paren {b \cap c}
      | r = \paren {a \cap b} \cap c
      | c = Intersection is Associative
}}
{{end-eqn}}
and $\RR$ fulfils {{Boolean-algebra-axiom|2|2}}.
{{qed|lemma}}
;{{Boolean-algebra-axiom|2|3|nolink}}:
We have:
{{begin-eqn}}
{{eqn | q = \forall a, b \in \RR
      | l = b \cup \paren {b \cap a}
      | r = b
      | c = Union Absorbs Intersection
}}
{{eqn | ll= \leadsto
      | l = \paren {a \cap b} \cup b
      | r = b
      | c = Intersection is Commutative, Union is Commutative
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | q = \forall a, b \in \RR
      | l = b \cap \paren {b \cup a}
      | r = b
      | c = Intersection Absorbs Union
}}
{{eqn | ll= \leadsto
      | l = \paren {a \cup b} \cap b
      | r = b
      | c = Intersection is Commutative, Union is Commutative
}}
{{end-eqn}}
and $\RR$ fulfils {{Boolean-algebra-axiom|2|3}}.
{{qed|lemma}}
;{{Boolean-algebra-axiom|2|4|nolink}}:
We have:
{{begin-eqn}}
{{eqn | q = \forall a, b, c \in \RR
      | l = a \cap \paren {b \cup c}
      | r = \paren {a \cap b} \cup \paren {a \cap c}
      | c = Intersection Distributes over Union
}}
{{eqn | l = a \cup \paren {b \cap c}
      | r = \paren {a \cup b} \cap \paren {a \cup c}
      | c = Union Distributes over Intersection
}}
{{end-eqn}}
and $\RR$ fulfils {{Boolean-algebra-axiom|2|4}}.
{{qed|lemma}}
;{{Boolean-algebra-axiom|2|5|nolink}}:
{{begin-eqn}}
{{eqn | q = \forall a, b \in \RR
      | l = \paren {a \cap \relcomp S a} \cup b
      | r = \O \cup b
      | c = Intersection with Relative Complement is Empty
}}
{{eqn | r = b
      | c = Union with Empty Set
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | q = \forall a, b \in \RR
      | l = \paren {a \cup \relcomp S a} \cap b
      | r = S \cap b
      | c = Union with Relative Complement and note 
}}
{{eqn | r = b
      | c = Intersection with Subset is Subset
}}
{{end-eqn}}
and $\RR$ fulfils {{Boolean-algebra-axiom|2|5}}.
{{qed|lemma}}
All of the Boolean algebra axioms (formulation $2$) are fulfilled.
Hence the result.
{{qed}}