# Peano axioms: prove that there is no natural number between n and sucessor of n

Let $$m \in \mathbb{N}$$. Show that $$\nexists n \in \mathbb{N}$$ such that $$m < n < s(m)$$, where $$s$$ is the successor function.

Here's my proof using only the Peano Axioms I was introduced. I'd appreciate someone to check my work.

Let's prove it by contradiction. Suppose that $$\exists n \in \mathbb{N}$$ such that (1) $$m < n$$ and (2) $$n < s(m)$$.

From the definition of order we have that:

(1) implies that $$\exists c \in \mathbb{N} \setminus \{0\}$$ such that $$m + c = n$$.

(2) imples that $$\exists c' \in \mathbb{N} \setminus \{0\}$$ such that $$n + c' = s(m)$$.

Therefore \begin{align*} n + c' + c &= s(m) + c\\ &= s(m + c)\\ &= s(n)\\ \end{align*}

Since $$c' \neq 0$$ and $$c \neq 0$$, we know that $$c' + c \neq 0$$ and therefore $$\exists k \in \mathbb{N}$$ such that $$s(k) = c' + c$$. Hence:

\begin{align*} n + s(k) &= s(n)\\ s(n + k) &= s(n) \rightarrow n + k = n \rightarrow k = 0 \end{align*}

Since we've concluded that $$k=0$$, we have that $$s(k) = s(0) = 1 = c + c'$$, where follows that $$c$$ or $$c'$$ needs to be equals to $$0$$, and that is a contradiction.

I also have one more question... After defining what a function is, can't we conclude that $$s : \mathbb{N} \setminus \{0\} \rightarrow \mathbb{N}$$ is actually $$s(n) = n + 1$$ ?

If so, using that the proof would've been more immediate...

Any kind of comments and critics are highly appreciated! Thank you!

Your proof doesn't work as it is, because from "$$n+k = n$$" you cannot get "$$k = 0$$" just like that. You need induction in order to prove this cancellation fact. Either that or you just use induction to directly prove the original desired theorem.