Strictly Precedes is Strict Ordering

Theorem
Let $\left({S, \preceq}\right)$ be a poset.

Let $\prec$ be the relation on $S$ defined as:
 * $a \prec b \iff a \ne b \land a \preceq b$

That is, $a \prec b$ iff $a$ strictly precedes $b$.

Then:
 * $a \preceq b \iff a = b \lor a \prec b$

and $\prec$ is a strict ordering on $S$.

Also note:
 * $a \prec b, b \preceq c \implies a \prec c$
 * $a \preceq b, b \prec c \implies a \prec c$

Proof
We are given that $\left({S, \preceq}\right)$ is a poset.


 * Antireflexive: Follows immediately:


 * $\forall a \in S: a = a \implies a \not \prec a$


 * Transitive:

Suppose $a, b, c \in S$ such that $a \preceq b, b \preceq c$.

$a \preceq b \land b \preceq c \implies a \preceq c$ from transitivity of $\preceq$.

Now suppose $a \preceq b, b \preceq c, a = c$. Then $a \preceq b$ but because of the antisymmetry of $\preceq$, it is not then possible for $b \preceq c$. So a condition for transitivity to be violated will not arise in this circumstance.

If either $a = b$ or $b = c$, then $a \not \prec b$ and $b \not \prec c$ by antireflexivity of $\prec$.

So $\prec$ is shown to be transitive.


 * Asymmetric:

A direct application of Relation Antireflexive and Transitive therefore Asymmetric.

Hence the result.


 * Let $a \preceq b$.

Either $a = b$ or $a \preceq b$ by the Law of Excluded Middle.

If $a = b$ we are done.

Suppose $a \preceq b$. Then it follows that $a \prec b$.

Thus $a \preceq b \implies a = b \lor a \prec b$.


 * Now let $a = b \lor a \prec b$.

If $a = b$ then $a \preceq b$ by the reflexivity of $\preceq$.

If $a \prec b$ then $a \preceq b$ by definition.