User:Alecscooper/Sandbox

Let $\left({S, *, \circ}\right)$ be a semiring, all of whose elements of $S$ are cancellable for $*$ with at either a right identity or a left identity (or both) for the distributor.

We will demonstrate the case where $\left({S, *, \circ}\right)$ has a right identity by expanding $\left({x * y}\right) \circ \left({r * r}\right)$ using the distributive law in two ways:

	\(\displaystyle \)	\(\displaystyle \left({ x * y }\right) \circ \left({ r * r }\right)\)	\(=\)	\(\displaystyle \left({ \left({ x * y }\right) \circ r }\right) * \left({ \left({ x * y }\right) \circ r }\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({ x * y }\right) * \left({ x * y }\right)\)	\(\displaystyle \)	Since $r$ is a right identity
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x * \left({ y * x }\right) * y\)	\(\displaystyle \)	Since $*$ is associative

But we also have

	\(\displaystyle \)	\(\displaystyle \left({ x * y }\right) \circ \left({ r * r }\right)\)	\(=\)	\(\displaystyle \left({ x \circ \left({ r * r }\right) }\right) * \left({ y \circ \left({ r * r }\right) }\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({ \left({ x \circ r }\right) * \left({ x \circ r }\right) }\right) * \left({ \left({ y \circ r }\right) * \left({ y \circ r }\right) }\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({ x * x }\right) * \left({ y * y }\right)\)	\(\displaystyle \)	Since $r$ is a right identity
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x * \left({ x * y }\right) * y\)	\(\displaystyle \)	Since $*$ is associative

Thus we have $x * \left({ x * y }\right) * y = x * \left({ y * x }\right) * y$.

But our semiring is cancellable, so this simplifies to $x * y = y * x$, which is exactly what we wanted to show.

Similarly, if $r$ was a left identity, then the above would follow similarly by expanding $\left({ r * r }\right) \circ \left({ x * y }\right)$.

$\blacksquare$