(related to Proposition: Uniqueness Of Predecessors Of Natural Numbers)
Let \(M\) be a subset of all natural numbers, which contains the number \(0\) and all \(x\neq 0\), which have at least one predecessor \(u\), i.e. for which the equation \(x=u^+\) holds.
We first show that \(M\) really contains not just the number \(0\), but more elements. This is the case, since \(0\in M\) and, by Peano axioms P1 and P2 and construction of \(M\), we have also \(0^+\in M\).
Note that, again according to the Peano axiom P2, for each \(x\in M\) with \(x\neq 0\) there exists a natural number \(x^+\) with \(x^+=(u^+)^+\), which is the unique successor of every \(x\in M\). By construction of \(M\), we have \(x^+\in M\). Since \(0\in M\), we can apply the principle of induction P5, and conclude that \(M\) is not only a subset of natural numbers, but \(M\) contains all natural numbers \(\mathbb N\), in particular, it contains \(u\in M\).
We have just shown that every \(x\neq 0\), \(x\in M\) has at least one predecessor \(u\in M\). By Peano axiom P4 it follows, that there is only one such predecessor, thus \(u\) is unique.