If $n,m\in \mathbb{Z}$ with $mWhat is the property in the title called? I have been trying to search for it in websites, but can't find anywhere specific. I learnt that

If $n,m\in \mathbb{Z}$ with $m<n$, then $m\leq n-1$.

I was not sure, if this is an axiom or a theorem, please let me know.
But anyway, I tried to prove this claim. It turned out that it is indeed equivalent to:

If $n\in \mathbb{Z}$, then there is no $m\in \mathbb{Z}$ such that $n-1<m<n$.

which can be proven by means of Well-ordering principle. To see the equivalence, we have
\begin{align}
\forall n\in \mathbb{Z}\neg(\exists m\in \mathbb{Z}:n-1<m<n)\\
\iff \forall n\in \mathbb{Z}\forall m\in \mathbb{Z}:(\neg(m<n)\vee (m\leq n-1))\\
\iff \forall n\in \mathbb{Z}\forall m\in \mathbb{Z}:(m<n \implies m\leq n-1)
\end{align}
Am I on the right track?
 A: What is the context? What axioms are you using? There is one context where one starts with axioms for $\mathbb{R}$ (as a complete ordered field) instead of constructing it from Dedekind cuts of $\mathbb{Q}$, and then defines $\mathbb{N}$ as the smallest set containing $1$ and having the property that if $x \in \mathbb{N}$ then $x+1 \in \mathbb{N}$, too. In this "advanced calculus lite" version this would be a theorem, proved by using the theorem/fact that if a set $A$ is a subset of $\mathbb{N}$ and also satisfies the conditions ($1\in A$ and $(x\in A) \Rightarrow (x+1 \in A)$) then $A=\mathbb{N}$. You would start by checking that the set $A = \{1\} \cup ([2,\infty) \cap \mathbb{N})$ is $\mathbb{N}$ because $1+1=2$ and for any $x \in [2,\infty) \cap \mathbb{N}$ you also have $x+1 \in [2,\infty)\cap \mathbb{N}$. Then you would prove by similar type of reasoning that for every $n \in \mathbb{N}$ you have $(n,n+1)\cap \mathbb{N}=\emptyset$. Let $A$ be the set of numbers $n\in \mathbb{N}$ satisfying this, which we know satisfies $1 \in A$. Now show that if $n \in A$ then also $n+1\in A$ by considering $B = ([1,n+1]\cup[n+2,\infty))\cap \mathbb{N}$ (and using the fact that $(n,n+1)\cap B=\emptyset$) and showing it equals $\mathbb{N}$.
Of course if you start with the construction of $\mathbb{N}$ from set-theoretic foundations, including the axiom of infinity, then the proof would be different.
