# Equivalence of Definitions of Semiring of Sets

## Theorem

The following definitions of the concept of Semiring of Sets are equivalent:

### Definition 1

Let $\SS$ be a system of sets.

$\SS$ is a semiring of sets or semi-ring of sets if and only if $\SS$ satisfies the semiring of sets axioms:

 $(1)$ $:$ $\ds \O \in \SS$ $(2)$ $:$ $\cap$-stable $\ds \forall A, B \in \SS:$ $\ds A \cap B \in \SS$ $(3)$ $:$ $\ds \forall A, A_1 \in \SS : A_1 \subseteq A:$ $\exists n \in \N$ and pairwise disjoint sets $A_2, A_3, \ldots, A_n \in \SS : \ds A = \bigcup_{k \mathop = 1}^n A_k$

### Definition 2

Let $\SS$ be a system of sets.

$\SS$ is a semiring of sets or semi-ring of sets if and only if $\SS$ satisfies the semiring of sets axioms:

 $(1)$ $:$ $\ds \O \in \SS$ $(2)$ $:$ $\cap$-stable $\ds \forall A, B \in \SS:$ $\ds A \cap B \in \SS$ $(3')$ $:$ $\ds \forall A, B \in \SS:$ $\exists n \in \N$ and pairwise disjoint sets $A_1, A_2, A_3, \ldots, A_n \in \SS : \ds A \setminus B = \bigcup_{k \mathop = 1}^n A_k$

## Proof

### Definition 1 implies Definition 2

Let $\SS$ be a system of sets satisfying the axioms:

 $(1)$ $:$ $\ds \O \in \SS$ $(2)$ $:$ $\cap$-stable $\ds \forall A, B \in \SS:$ $\ds A \cap B \in \SS$ $(3)$ $:$ $\ds \forall A, A_1 \in \SS : A_1 \subseteq A:$ $\exists n \in \N$ and pairwise disjoint sets $A_2, A_3, \ldots, A_n \in \SS : \ds A = \bigcup_{k \mathop = 1}^n A_k$

It remains to be shown that $\SS$ satisfies the axiom

 $(3')$ $:$ $\ds \forall A, B \in \SS:$ $\exists n \in \N$ and pairwise disjoint sets $A_1, A_2, A_3, \ldots, A_n \in \SS : \ds A \setminus B = \bigcup_{k \mathop = 1}^n A_k$

Let $A, B \in \SS$.

Let $A_1 = A \cap B$.

By axiom $(2)$:

$A_1 \in \SS$
$A_1 \subseteq A$

By axiom $(3)$:

$\exists$ a finite sequence of pairwise disjoint sets $A_2, A_3, \ldots, A_n \in \SS : \ds A = \bigcup_{k \mathop = 1}^n A_k$

Then:

 $\ds A \setminus B$ $=$ $\ds A \setminus \paren {A \cap B}$ Set Difference with Intersection is Difference $\ds$ $=$ $\ds A \setminus A_1$ $\ds$ $=$ $\ds \paren {\bigcup_{k \mathop = 1}^n A_k} \setminus A_1$ $\ds$ $=$ $\ds \bigcup_{k \mathop = 1}^n \paren {A_k \setminus A_1}$ Set Difference is Right Distributive over Union $\ds$ $=$ $\ds \bigcup_{k \mathop = 2}^n \paren {A_k \setminus A_1}$ Set Difference with Self is Empty Set and Union with Empty Set $\ds$ $=$ $\ds \bigcup_{k \mathop = 2}^n A_k$ Set Difference with Disjoint Set

As $A$ and $B$ were arbitrary, then $\SS$ satisfies axiom $(3')$

The result follows

$\Box$

### Definition 2 implies Definition 1

Let $\SS$ be a system of sets satisfying the axioms:

 $(1)$ $:$ $\ds \O \in \SS$ $(2)$ $:$ $\cap$-stable $\ds \forall A, B \in \SS:$ $\ds A \cap B \in \SS$ $(3')$ $:$ $\ds \forall A, B \in \SS:$ $\exists n \in \N$ and pairwise disjoint sets $A_1, A_2, A_3, \ldots, A_n \in \SS : \ds A \setminus B = \bigcup_{k \mathop = 1}^n A_k$

It remains to be shown that $\SS$ satisfies the axiom

 $(3)$ $:$ $\ds \forall A, A_1 \in \SS : A_1 \subseteq A:$ $\exists n \in \N$ and pairwise disjoint sets $A_2, A_3, \ldots, A_n \in \SS : \ds A = \bigcup_{k \mathop = 1}^n A_k$

Let $A, B \in \SS : B \subseteq A$.

By axiom $(3)$:

$\exists$ a finite sequence of pairwise disjoint sets $A_1, A_2, \ldots, A_n \in \SS : \ds A \setminus B = \bigcup_{k \mathop = 1}^n A_k$.

Then $B$ is disjoint with each of the sets $A_k$.

Then:

 $\ds B \cup \bigcup_{k \mathop = 1}^n A_k$ $=$ $\ds \paren {A \setminus B} \cup B$ $\ds$ $=$ $\ds A \cup B$ Set Difference Union Second Set is Union $\ds$ $=$ $\ds A$ Union with Superset is Superset

As $A$ and $B$ were arbitrary, then $\SS$ satisfies axiom $(3)$

The result follows

$\blacksquare$