# Proof

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.

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### References

#### Bibliography

1. Landau, Edmund: "Grundlagen der Analysis", Heldermann Verlag, 2008