the background I'm allowed to deal with to solve this problem is as follows:
Definition of +: \begin{equation} m+0=m\quad \text{for all}\quad m \in \mathbb{N} \\ m+(k+1) = (m+k)+1 \end{equation} in conjunction with the induction property of $\mathbb{N}$. In a previous task I showed that $ 0+m=m \quad \forall \, m \in \mathbb{N}$.
Task: Prove $m+n=n+m$ by induction on $n$.
Now, it follows directly that $0+m=m=m+0$. With the assumption that $k+m=m+k$ for a $k\in\mathbb{N}$, it can be deduced: \begin{eqnarray} m+(k+1) & = & (m+k)+1\quad \text{by def. of}+\\ & = & (k+m) + 1\quad \text{with assumption}\\ & = & k + (m+1)\quad \text{by def. of}+. \end{eqnarray} Here, it looks to me that the true base step would be $m+1 = 1+m$, because then it would be possible to bring the proof straight to an end. But I tried hard to hack this last nut without any results. Is there anybody who could help me with this? Or maybe there is a completely different way to come up with the proof requested?
Thanks in advance for your help
Ok folks, @Eleven-Eleven figured it out. The definition of $\mathbb{N}$ that I use is: \begin{equation} \mathbb{N}\, \text{is the closure of the set} \{0\}\,\text{under}\,\it{successor}. \end{equation} This is, I think, basically the same definition as in the comment. Now let $m>0$ be the successor of $k$: \begin{equation} m=k+1. \end{equation} Then \begin{eqnarray} 1+m&=&1+(k+1)\\ &=&(1+k)+1\quad \text{by def. of}+\\ &=&m+1. \end{eqnarray} With this, starting from the last manipulation in my question: \begin{eqnarray} m+(k+1)&=&k+(m+1)\\ &=&k+(1+m)\\ &=&(k+1)+m. \end{eqnarray}