# Cancellable Semiring with Unity is Additive Semiring

## Theorem

Let $\struct {S, *, \circ}$ be a cancellable semiring with unity $1_S$.

Then the distributand $*$ is commutative.

That is to say, $\struct {S, *, \circ}$ is also an additive semiring.

## Proof

Let $\struct {S, *, \circ}$ be a semiring, all of whose elements of $S$ are cancellable for $*$.

We expand the expression $\paren {a * b} \circ \paren {c * d}$ using the distributive law in two ways:

 $\displaystyle \paren {a * b} \circ \paren {c * d}$ $=$ $\displaystyle \paren {\paren {a * b} \circ c} * \paren {\paren {a * b} \circ d}$ $\displaystyle$ $=$ $\displaystyle \paren {a \circ c} * \paren {b \circ c} * \paren {a \circ d} * \paren {b \circ d}$

 $\displaystyle \paren {a * b} \circ \paren {c * d}$ $=$ $\displaystyle \paren {a \circ \paren {c * d} } * \paren {b \circ \paren {c * d} }$ $\displaystyle$ $=$ $\displaystyle \paren {a \circ c} * \paren {a \circ d} * \paren {b \circ c} * \paren {b \circ d}$

So, by the fact that all elements of $\struct {S, *}$ are cancellable (and thus, a fortiori, are $a \circ c$ and $b \circ d$), we have:

 $\displaystyle \paren {a \circ c} * \paren {b \circ c} * \paren {a \circ d} * \paren {b \circ d}$ $=$ $\displaystyle \paren {a \circ c} * \paren {a \circ d} * \paren {b \circ c} * \paren {b \circ d}$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {b \circ c} * \paren {a \circ d}$ $=$ $\displaystyle \paren {a \circ d} * \paren {b \circ c}$

As this is true for all $a, b, c, d \in S$, it is true in particular if $c = d = 1_S$.

Thus it is clear that $b * a = a * b$, which is exactly to say that $*$ is commutative.

Hence the result, by definition of additive semiring.

$\blacksquare$