Exercise 5, § 3

On a set $E$ suppose three internal laws of composition are given: an addition (not necessarily associative nor commutative), a multiplication (not necessarily associative) and a law denoted by $\top$. Suppose that the multiplication has an identity element $e$, that the law $\top$ is right distributive with respect to multiplication and the law $\top$ is left distributive with respect to addition. Show that, if there exist $x, y, z$ such that $x \top z, y \top z$ and $(x + y) \top z$ are cancellable under multiplication, then $e + e = e$ (use Exercise 3(a)).