Cancellable Semiring: Distributand Commutative
Theorem
In a cancellable semiring $\left({S, *, \circ}\right)$, the distributand $*$ is always commutative.
Proof
Let $\left({S, *, \circ}\right)$ be a semiring, all of whose elements of $S$ are cancellable for $*$.
We expand the expression $\left({a * b}\right) \circ \left({c * d}\right)$ using the distributive law in two ways:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left({a * b}\right) \circ \left({c * d}\right)\) | \(=\) | \(\displaystyle \left({\left({a * b}\right) \circ c}\right) * \left({\left({a * b}\right) \circ d}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({a \circ c}\right) * \left({b \circ c}\right) * \left({a \circ d}\right) * \left({b \circ d}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left({a * b}\right) \circ \left({c * d}\right)\) | \(=\) | \(\displaystyle \left({a \circ \left({c * d}\right)}\right) * \left({b \circ \left({c * d}\right)}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({a \circ c}\right) * \left({a \circ d}\right) * \left({b \circ c}\right) * \left({b \circ d}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
So, by the fact that all elements of $\left({S, *}\right)$ are cancellable (and thus are $a \circ c$ and $b \circ d$), we have:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left({a \circ c}\right) * \left({b \circ c}\right) * \left({a \circ d}\right) * \left({b \circ d}\right)\) | \(=\) | \(\displaystyle \left({a \circ c}\right) * \left({a \circ d}\right) * \left({b \circ c}\right) * \left({b \circ d}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle \left({b \circ c}\right) * \left({a \circ d}\right)\) | \(=\) | \(\displaystyle \left({a \circ d}\right) * \left({b \circ c}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
As this is true $\forall a, b, c, d \in \left({S, *, \circ}\right)$, it follows that for the distributive law to work, then $*$ must be commutative.
$\blacksquare$
Are all elements of $S$ representable in the form $a \circ d$?
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