Order-Extension Principle

Theorem
Let $S$ be a set.

Let $\preceq$ be an ordering on $S$.

Then there exists a total ordering $\le$ on $S$ such that:
 * $\forall a, b \in S: \left({a \preceq b \implies a \le b}\right)$

Remarks
As shown in Proof 2, the order-extension principle is weaker than the Boolean Prime Ideal Theorem (BPI).

It is known that it is in fact strictly weaker than BPI.

However, it cannot be proved in Zermel-Fraenkel set theory without the Axiom of Choice. In fact it is known to be strictly stronger than the Ordering Principle.