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!
