User:Jshflynn/Sandbox2

Theorem
The following definitions of a rectangular band are equivalent:

Cartesian Product implies xyx = x Identity
Let $\left({S \times T, \circ}\right)$ be a rectangular band.

That is, let $\left({S \times T, \circ}\right)$ be an algebraic structure such that:


 * $\forall \left({a, b}\right), \left({c, d}\right) \in S \times T: \left({a, b}\right) \circ \left({c, d}\right) = \left({a, d}\right)$

Let $\left({a, b}\right), \left({c, d}\right), \left({e, f}\right) \in S \times T$.

Proving $\circ$ is Idempotent

 * $\left({a, b}\right) \circ \left({a, b}\right) = \left({a, b}\right)$

Thus it follows directly from the definition of $\circ$ that $\circ$ is idempotent.

Proving $\circ$ is Associative
By the definition of $\circ$ :

Then for the other way in which it could have been bracketed:

Hence $\circ$ is associative.

Proving $\circ$ Satisfies the Identity
By the definition of $\circ$ :

By $\circ$ being associative the following holds as well:


 * $\left({a, b}\right) \circ \left({\left({c, d}\right) \circ \left({a, b}\right)}\right) = \left({a, b}\right)$

Hence $\circ$ satisfies the identity:


 * $\forall x, y \in S \times T: x \circ y \circ x = x$