Equivalence of Definitions of Ring of Sets

Definition 1 implies Definition 2
Let $\RR$ be a system of sets such that for all $A, B \in \RR$:
 * $(\text {RS} 1_1): \quad \RR \ne \O$
 * $(\text {RS} 2_1): \quad A \cap B \in \RR$
 * $(\text {RS} 3_1): \quad A \symdif B \in \RR$

As $\RR$ is non-empty, there exists some $A \in \RR$.

From Symmetric Difference with Self is Empty Set:
 * $A \symdif A = \O$

By hypothesis $A \symdif A \in \RR$ and so $\O \in \RR$.

Thus criterion $(\text {RS} 1_2)$ is fulfilled.

From Closure of Intersection and Symmetric Difference imply Closure of Set Difference it follows that criterion $(\text {RS} 2_2)$ is fulfilled.

From Closure of Intersection and Symmetric Difference imply Closure of Union it follows that criterion $(\text {RS} 3_2)$ is fulfilled.

Definition 2 implies Definition 1
Let $\RR$ be a system of sets such that for all $A, B \in \RR$:
 * $(\text {RS} 1_2): \quad \O \in \RR$
 * $(\text {RS} 2_2): \quad A \setminus B \in \RR$
 * $(\text {RS} 3_2): \quad A \cup B \in \RR$

We have that $\O \in \RR$ and so $\RR$ is non-empty.

Thus criterion $(\text {RS} 1_1)$ is fulfilled.

By hypothesis, $\RR$ is closed under $\setminus$ and $\cup$.

Thus:
 * $\forall A, B \in \RR: \paren {A \setminus B} \cup \paren {B \setminus A} \in \RR$

But by the definition of symmetric difference:
 * $A \symdif B := \paren {A \setminus B} \cup \paren {B \setminus A}$

Thus:
 * $\forall A, B \in \RR: A \symdif B \in \RR$

and so $\RR$ is closed under symmetric difference.

Thus criterion $(\text {RS} 3_1)$ is fulfilled.

From Union minus Symmetric Difference equals Intersection:
 * $\forall A, B \in \RR: \paren {A \cup B} \setminus \paren {A \symdif B} = A \cap B$

Thus $\RR$ is closed under set intersection.

Thus criterion $(\text {RS} 2_1)$ is fulfilled.

Definition 2 iff Definition 3
Let $\RR$ be a system of sets such that for all $A, B \in \RR$:
 * $(\text {RS} 1_2): \quad \O \in \RR$
 * $(\text {RS} 2_2): \quad A \setminus B \in \RR$
 * $(\text {RS} 3_2): \quad A \cup B \in \RR$

Criteria $(\text {RS} 1_3)$ and $(\text {RS} 2_3)$ are fulfilled immediately.

Consider $A, B \in \RR: A \cap B = \O$.

Then as $A, B \in \RR$ it follows by $(\text {RS} 3_2)$ that $A \cup B \in \RR$ and so $(\text {RS} 3_3)$ is fulfilled.

Now let $\RR$ be a system of sets such that for all $A, B \in \RR$:
 * $(\text {RS} 1_3): \quad \O \in \RR$
 * $(\text {RS} 2_3): \quad A \setminus B \in \RR$
 * $(\text {RS} 3_3): \quad A \cap B = \O \implies A \cup B \in \RR$

Again, criteria $(\text {RS} 1_2)$ and $(\text {RS} 2_2)$ are fulfilled immediately.

Let $A, B \in \RR$.

Then from Set Difference Union Second Set is Union:
 * $A \cup B = \paren {A \setminus B} \cup B$

From Set Difference Intersection with Second Set is Empty Set:
 * $\paren {A \setminus B} \cap B = \O$

Thus from $(\text {RS} 3_3)$:
 * $A \cup B = \paren {A \setminus B} \cup B \in \RR$