Let $m,n \in\mathbb Z$. If $m \le n \le m$ then $m = n$. I an learning proofs with $\mathbb{N}$. 
Here are my axioms:

a)If $m,n \in\mathbb N$ then $m + n \in\mathbb N$

b)If $m,n \in\mathbb N$ then $mn \in\mathbb N$ 

c) $0 \notin\ \mathbb N$

d) For every $m \in\mathbb Z$, we have $m \in\mathbb N$ or $m = 0$ or $-m \in\mathbb N$

Definition: $m &gt; n  \Leftrightarrow m - n \in\mathbb N$.

By the way, I would greatly appreciate if someone could please explain to me why this is. My strategy is to use a contradiction.

Proposition 1: (that I have proven)

For $m \in\mathbb Z$, one and only one of the following is true: $m \in\mathbb N$, $-m \in\mathbb N$, $m = 0$.

I have proven $1 \in\mathbb N$ and: Let $m,n,p \in\mathbb Z$. If $m &lt; n$ and $n &lt; p$ then $m &lt; p$ by deriving $p - m = (p - n) + (n - m) \in\mathbb N$.

I have also proven: For each $n \in\mathbb N$ there exists $m \in\mathbb N$ such that $m &gt; n$

Proposition 2:

Let $m,n \in\mathbb Z$. If $m \le n \le m$ then $m = n$. 

I get confused because of the equality within the $\leq$. How could I approach this? By separating $&lt;$ and $=$ ?

Any hints would be greatly appreciated.
First proposition


If $m \in \mathbb N$ and $-m \in \mathbb N$ then $ 0 \in \mathbb N$, absurd. 
If $m \in \mathbb N$ and $ m = 0$, absurd. 
Similar to $-m$. 


Therefore it folows the affirmation. 

Second proposition

If $m \leq n \leq m$ then $m - n \in \mathbb Z$ and by the first proposition only one of those $3$ holds, then if $m - n \in \mathbb N$ then $m&gt;n$ by definition, which is a contradiction. Similar if $n-m \in \mathbb N$. 

Thus the only possibility is $m - n = 0$.
Suppose $m \ne n$. That means: $m &lt; n &lt; m$. Then by definition, $m−n∈N$ and $n-m∈N$. By axiom (a): $0=m-n+n-m∈N$. Came to contradiction by axiom (c), so $m$ is indeed equal to $n$.