Well-Ordering Principle: $\exists m \in A[\forall n \in A (m \in n \vee m = n) ]$ for all $A \subseteq w$ where $w$ is the set of natural numbers and $A \neq \phi$.
Base Case: $A_{1} = \{ e_{1} \}$ is trivially true.
Induction Hypothesis: Let $A_{n} = \{ e_{1}, ..., e_{n} \}$ which has a smallest element, $e_{l}$.
Suppose $A_{n+1} = \{ e_{1}, ..., e_{n+1} \}$ where $A_{n+1} = A_{n} \cup \{e_{n+1}\} $.
By Trichotomy Law, 1) $e_{l} \in e_{n+1}$, or 2) $e_{l} = e_{n+1}$ or 3) $e_{n+1} \in e_{l}$.
If 1), $e_{l}$ continues to be the smallest element of $A_{n+1}$.
If 2), $A_{n} = A_{n+1}$ where $e_{l}$ is the smallest element.
If 3), as $\in$ is transitive, $e_{n+1} \in e_{i}$ for $\forall e_{i} \in A_{n} $ while being equal to itself. Therefore $e_{n+1}$ is the smallest of $A_{n+1}$.
Therefore, all subsets of $w$ has a smallest element. $\Box$