Does a proof using the well-ordering principle need a base case?

For proofs by well-ordering principle the general template is to consider the negation of some predicate $P(n)$. Then assume the set of all elements that fulfill $\lnot P(n)$, i.e.

$\qquad N = \{ n \mid \lnot P(n) \}$

has a smallest element according to WOP, say $m = \min N$, and if we manage to prove that there is another element $m' \in N$ that is less than $m$ that also negates then we contradict our assumption that $m$ is the smallest element.

My question is that should we be proving some sort of a base case as well for the above mentioned template. As in proving that for some base case, $P(\mathrm{base})$ is true?

In the Wiki page,


In mathematics, the well-ordering principle states that every non-empty set of positive integers contains a least element.

Pay attention to the word non-empty.

Consider the negation of P(n). Before you assume the set that satisfying P(n) is false has a smallest element m, you already assume that the set is non-empty. You prove that there is another element that is less than m. This contradicts the assumption that the set is non-empty. Thus, the set is empty which means P(n) is true.

StackExchange Computer Science Q#14082, answer score: 4