EDB — 1XB

[1XB](Solved on 2022-11-03)

(N1)

There is a number \(0∈ ℕ\).

(N2)

There is a function \(S:ℕ → ℕ\) (called "successor"), such that ¹

(N3)

\(S(x)≠ 0\) for each \(x∈ ℕ\) and

(N4)

\(S\) is injective, that is, \(x≠ y\) implies \(S(x)≠ S(y)\).

(N5)

If \(U\) is a subset of \(ℕ\) such that: \(0∈ U\) and \(∀ x, x ∈ U⇒ S(x)∈ U\) , then \(U=ℕ\).

We will often write \(Sn\) instead \(S(n)\) to ease notations.