Proof
(related to Proposition: WellOrdering Principle of Natural Numbers)
We will prove the proposition for the poset $(\mathbb N,\le).$ An analogous proof is possible for the strictlyordered set $(\mathbb N, < ),$ in which we have to replace the order relation "$\le$" by the order relation "$<$" and the word minimum by the word minimal element.
 Let \(M\) be a nonempty subset of natural numbers \(M\subseteq\mathbb N.\)
 Since $M$ is nonempty, let \(n_0\in M\) be its element.
 We will construct a minimum as follows:
 If there is no smaller element \(m \le n_0\) then we set \(m_0:=n_0\), since it is the minimum we were looking for.
 Otherwise there exists an $n_1\in M$ with \(n_1 < n_0.\)
 We ask again if there is no smaller element \(m \le n_1\).
 If so, then we set \(m_0:=n_1\), otherwise we find an $n_2\in M$ with \(n_2 < n_1.\) and repeat the same search again.
 Our searching process will be repeated at most \(n_0  m_0+1\) times until we finally find the smallest element of \(M\).
 This shows that $M$ contains a minimum, which is by construction unique.
 We have shown that $(\mathbb N,\le)$ is a wellordered set.
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References
Bibliography
 Kramer Jürg, von Pippich, AnnaMaria: "Von den natürlichen Zahlen zu den Quaternionen", SpringerSpektrum, 2013