Equivalence of Definitions of Ring of Sets

Theorem
The following definitions of ring of sets are equivalent:

Definition 1 implies Definition 2
Let $\mathcal R$ be a system of sets such that for all $A, B \in \mathcal R$:
 * $(RS \, 1_1): \quad \mathcal R \ne \varnothing$
 * $(RS \, 2_1): \quad A \cap B \in \mathcal R$
 * $(RS \, 3_1): \quad A * B \in \mathcal R$

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

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

By hypothesis $A * A \in \mathcal R$ and so $\varnothing \in \mathcal R$.

Thus criterion $(RS \, 1_2)$ is fulfilled.

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

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

Definition 2 implies Definition 1
Let $\mathcal R$ be a system of sets such that for all $A, B \in \mathcal R$:
 * $(RS \, 1_2): \quad \varnothing \in \mathcal R$
 * $(RS \, 2_2): \quad A \setminus B \in \mathcal R$
 * $(RS \, 3_2): \quad A \cup B \in \mathcal R$

We have that $\varnothing \in \mathcal R$ and so $\mathcal R$ is non-empty.

Thus criterion $(RS \, 1_1)$ is fulfilled.

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

Thus:
 * $\forall A, B \in \mathcal R: \left({A \setminus B}\right) \cup \left({B \setminus A}\right) \in \mathcal R$

But by the definition of symmetric difference:
 * $A * B := \left({A \setminus B}\right) \cup \left({B \setminus A}\right)$

Thus:
 * $\forall A, B \in \mathcal R: A * B \in \mathcal R$

and so $\mathcal R$ is closed under symmetric difference.

Thus criterion $(RS \, 3_1)$ is fulfilled.

From Union minus Symmetric Difference equals Intersection:
 * $\forall A, B \in \mathcal R: \left({A \cup B}\right) \setminus \left({A * B}\right) = A \cap B$

Thus $\mathcal R$ is closed under set intersection.

Thus criterion $(RS \, 2_1)$ is fulfilled.

Definition 2 iff Definition 3
Let $\mathcal R$ be a system of sets such that for all $A, B \in \mathcal R$:
 * $(RS \, 1_2): \quad \varnothing \in \mathcal R$
 * $(RS \, 2_2): \quad A \setminus B \in \mathcal R$
 * $(RS \, 3_2): \quad A \cup B \in \mathcal R$

Criteria $(RS \, 1_3)$ and $(RS \, 2_3)$ are fulfilled immediately.

Consider $A, B \in \mathcal R: A \cap B = \varnothing$.

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

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

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

Let $A, B \in \mathcal R$.

Then from Set Difference Union Second Set is Union:
 * $A \cup B = \left({A \setminus B}\right) \cup B$

From Set Difference Intersection with Second Set is Empty Set:
 * $\left({A \setminus B}\right) \cap B = \varnothing$

Thus from $(RS \, 3_3)$:
 * $A \cup B = \left({A \setminus B}\right) \cup B \in \mathcal R$