Exercise 2, § 3

Let $\perp$ be an internal law on $E$ which is doubly distributive with respect to the associative internal law $\top$; show that, if $x \perp x'$ and $y \perp y'$ are cancellable under the law $\top$, $x \perp y'$ is permutable with $y \perp x'$ under the law $\top$ (calculate the composition $(x \top y) \perp (x' \top y')$ in two different ways). In particular, if the law $\perp$ has an identity element, two cancellable elements under the law $\top$ are permutable under this law; if all the elements of $E$ are cancellable under the law $\top$, this law is commutative.