Well-Founded Relation is not necessarily Ordering

Theorem
Let $\struct {S, \RR}$ be a relational structure.

Let $\RR$ be a well-founded relation on $S$.

Then it is not necessarily the case that $\RR$ is also either an ordering or a strict ordering.

Proof
Proof by Counterexample:

Let $P$ be the set of all polynomials over $\R$ in one variable with real coefficients.

Let $\DD$ be a relation on $P$ defined as:
 * $\forall p_0, p_1 \in P: \tuple {p_0, p_1} \in \DD$ $p_0$ is the derivative of $p_1$.

From Differentiation of Polynomials induces Well-Founded Relation, we have that $\DD$ is a well-founded relation on $P$.

Let $P_a$ be the polynomial defined as:
 * $P_a = x^3$

Then the derivative of $P_a$ $x$ is:
 * $P_a' = \dfrac \d {\d x} x^3 = 3 x^2$

Then the derivative of $P_a$ $x$ is:
 * $P_a'' = \dfrac \d {\d x} 3 x^2 = 6 x$

So we have that:
 * $\tuple {P_a', P_a} \in \DD$

and:
 * $\tuple {P_a'', P_a'} \in \DD$

but it is not the case that $\tuple {P_a'', P_a} \in \DD$.

That is, $\DD$ is not transitive.

It follows that $\DD$ is neither an ordering nor a strict ordering.