Basically the title. By $s$ I mean the function $s: \mathbb{N} \to \mathbb{N}\setminus\{1\}$ defined by $s(n) = n + 1$. I only have induction, so I proved that $s(n) \neq n$, for the base of induction, and when I suppose $s^k(n) \neq n$, I can reach $s^{k+1}(n) \neq s(n)$, but in the left side I can't just put $n$. I got that $s^{k+1}(n) \neq s(n)$
$s(n) \neq n$
How can I conclude that $s^{k+1}(n) \neq n?$, since $\neq$ is not transitive? I guess I'm missing something really simple but I'm not seeing it.
We claim that $\forall k\in\mathbb{Z}^+, s^k(n)=n+k$. The proof is as follows:
Base case: $\forall n\in\mathbb{N}, s^1(n)=s(n)=n+1$
Induction hypothesis: $\forall n\in\mathbb{N}, s^k(n)=n+k$ where $k\in\mathbb{Z}^+$.
Inductive step: We claim that $\forall n\in\mathbb{N}, s^{k+1}(n)=n+k+1$.
$\forall n\in\mathbb{N}, s^{k+1}(n)=s^k(s(n))=s^k(n+1)=n+1+k$ by our induction hypothesis.
$\therefore\forall n\in\mathbb{N}, s^{k+1}(n)=n+k+1$. This completes the induction.
Now from our claim, we have that $\forall k, n\in\mathbb{N}, s^k(n)=n+k\neq n$.