Zero Strictly Precedes One

Theorem
Let $\left({S, \circ, \preceq}\right)$ be a naturally ordered semigroup.

Let $0$ be the zero of $S$.

Let $1$ be the one of $S$.

Then:


 * $0 \prec 1$

Proof
This follows directly from the definition of $\prec$.

First note that:


 * $\forall a \in S: 0 \preceq a$

from the definition of zero.

Next, from the definition of one:


 * $0 \ne 1$

Thus: