Equivalence of Definitions of Semiring of Sets/Definition 2 implies Definition 1

Theorem

Let $\SS$ be a system of sets satisfying 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 $A_1, A_2, A_3, \ldots, A_n \in \SS : \ds A \setminus B = \bigcup_{k \mathop = 1}^n A_k$

Then $\SS$ 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 $A_2, A_3, \ldots, A_n \in \SS : \ds A = \bigcup_{k \mathop = 1}^n A_k$

Proof

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

$\blacksquare$