This proof of peano's systems seemed so easy, perhaps i'm wrong?

Question

Let $P=\{1,2,3,4\}$, where the succesor of a number is $S(n)=n+1$ and $S(4)=1.$

This last one means that 1 is the succesor of another number, hence this can't be a Peano's system.

Do I have to prove it doesn't meet the other requirements?

Answer 1

Do I have to prove it doesn't meet the other requirements?

No. It actually meets the other requirements:

1. $1 \in P$
2. $S: P\to P$
3. $S$ is injective
4. $\forall A\subset P: [1 \in A \land \forall x\in A:[S(x)\in A] \implies A=P]$

Assuming:

$P=\{ 1,2, 3,4\}$

$S(1)=2$

$S(2)=3$

$S(3)=4$

$S(4)=1$