Knowing that, in $\mathbb{N}$, $m+n \neq n$, show that if $n\leq m$ and $m\leq n$, then $m=n$. 
Given the theorem that, if $m,n \in \mathbb{N}$ (the "counting numbers" without zero), then $m+n \neq n$.
Prove the reflexive property in $\mathbb{N}$; that is, if $n \leq m$ and $m \leq n$, then $m = n$.

My go was to try to derive a contradiction from the first theorem, but I was not able to relate them. It is as follows:
$n \leq m$ by definition means ($n < m$) or ($n = m$),
this means $ \exists \ k \in \mathbb{N}$ such that $ m = n + k$ or n = m
Then the same definition is for $\ n \leq m $.
So I want to show that the first case ($ m = n + k$) cannot be true, and then we will be left with $n = m$ as desired.
Any suggestion on how to show that (or possibly another way of deriving the contradiction)?
Thanks,
 A: Assumptions: $(1).\forall x,y\in \Bbb N\,(x+y\in \Bbb N)$.... $(2).$ $+$ is associative and commutative on $\Bbb N$.... $(3).  \forall x,y \in \Bbb N\,(x+y\ne x).$
Let $\Bbb N_0=\Bbb N\cup \{0\}$ where $\forall x \in \Bbb N_0\,(x+0=0+x=x).$ We easily verify that $+$ is associative and commutative on $\Bbb N_0.$
$(\bullet).\; 0\not \in \Bbb N.\;$ Proof: If $0\in \Bbb N,$ let $0=y\in \Bbb N.$ Then there exists some (any) $x\in \Bbb N$ such that $x=x+0=x+y,$ contrary to $(3).$
For $m,n\in \Bbb N$ we have $m\le n\iff \exists k\in \Bbb N_0\,(m+k=n).$
Let $m,n\in \Bbb N$ with $m\le n \le m.$ Let $k,k'\in \Bbb N_0$ such that $m+k=n$ and $n+k'=m.$ Then $$m+n=(n+k')+(m+k)=(m+n)+(k'+k).$$ Now with $m'=m+n$ and $n'=k'+k\in \Bbb N_0$ we have $m'\in \Bbb N$ and $m'+n'=m',$  which by $(3)$ implies that $k'+k=n'\not \in \Bbb N.$ So $k'+k=0.$
Now $(k'=0\implies n=n+0=n+k'=m),$ while $(k=0\implies m=m+0=m+k=n).$
But $k'\ne 0 \ne k \implies k',k\in \Bbb N\implies k'+k\in \Bbb N\implies k'+k\ne 0$ by $(\bullet),$  contrary to $k'+k=0.$ So this case does not occur.
