How do I prove that every ring is a semiring? Note: Take note that since I wrote this question approximately a year ago, I have taken a probability theory course.
This isn't a homework assignment. I'm trying to teach myself probability theory this summer and am currently going through the introduction on measure theory in Klenke's Probability Theory. 
I am trying to prove Theorem 1.12(ii). Klenke does go through some detail, but he does not complete the proof. The statement is:

Every $\sigma$-ring is a ring, and every ring is a semiring.

I have the following:

Let $\mathcal{A}$ be a $\sigma$-ring. Then $\mathcal{A}$ is obviously a ring. Since $\mathcal{A}$ is $\setminus$-closed, by Theorem 1.4., it is $\cap$-closed. Now let $A,B \in \mathcal{A}$. Consider $B \setminus A$...

I need to show that $B \setminus A$ is a finite union of mutually disjoint sets in $\mathcal{A}$. I've already seen the ProofWiki version of this; however, it uses a different definition. I was considering taking $B \setminus A = B \cap A^{C} = (B^C \cup A)^C$, but alas, $\mathcal{A}$ isn't closed under complements.
Any help is appreciated. (This is my first post here - hope I did this right!)
 A: I googled the same question and then was lead to your post here. I think I've figured out a full proof.
Before my written, I list both definitions of Ring and Semiring in Halmos' Measure Theory, that is,
A Ring(or Boolean Ring) of sets is a non empty class $R$ of sets such that if $$ E ∈ R \ and \ F ∈ R$$ then $$(E ∪ F) ∈ R \ and \ (E - F) ∈ R$$.
A Semiring of sets is a non empty class $P$ of Sets such that, if $$A, B ∈P$$ then $$(A ∩ B) ∈ P \ and \ (A - B)∈ P_{Σf}$$ where $P_{Σf} =$ {$A = \left(\bigcup_{i=1}^n A_i\right)| A_i ∈ P, A_i ∩ A_j = \emptyset \ for \ i≠j, i,j ∈$ {$i,2,3,...,n$}, $n \ is \ finite$}(Please note $n$ is finite).
Now begin my proof that is if $C$ is a Ring then try to show it is a Semiring as well.
Let $G_1$ = {$A' ∈ C| (A' ∩ B) ∈ C, (A' - B) ∈ C_{∑f}, \forall B ∈ C$} then it is clear $G_1 \subset C$.
Let $\forall A' ∈ C$.
Since $\forall B ∈ C$ $=>$ $(A'-B) ∈ C$ $=>$ $\exists D_{A'} ∈ C$ such that $(A'-B) = D_{A'}$. Since $C \subset C_{∑f}$ then $D_{A'} ∈ C_{∑f}$(Actually $D_{A'} = D_{A'} ∪ \emptyset$ that made $D_{A'} ∈ C_{∑f}$). Here we are on halfway to show $C \subset G_1$.
Next since $(A'∩B) = A'-(A'-B)$ and we still have $(A'-B) ∈ C$ due to $A', B ∈ C$, then $(A'∩B) ∈ C$.
So $A' ∈ G_1$.
Since $\forall A'$, then $$C \subset G_1$$.
Combined, $C = G_1$. Since $G_1$ is a Semiring, vacuously, $C$ can't tell a lie now^_^
