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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

C. Moos
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    if you are proving commutativity you can't start with $m+0=m=0+m$. You're base case has to be a member of you index set, in this case $\mathbb{N}$. I believe you are right to start with $n=1$. Maybe you just say, by the definition of a number, $m=\sum_{k=1}^m{1}$? Then $$m+1=\sum_{k+1}^{m}{1}+1=\sum_{k+1}^{m+1}{1}=1+\sum_{k+1}^{m}{1}=1+m$$ – Eleven-Eleven May 07 '14 at 21:31
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    Also, see http://math.stackexchange.com/questions/297609/how-to-prove-that-is-commutative-on-the-natural-numbers?rq=1 – Eleven-Eleven May 07 '14 at 21:44

2 Answers2

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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}

C. Moos
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    '1+m===1+(k+1)(1+k)+1by def. of+m+1.' I disagree; you are assuming that k+1=1+k which is what you are trying to prove. you need an inductive step. –  Nov 09 '15 at 13:59
  • @user288453 This solution seems valid to me. By the definition of +, m+(k+1)=(m+k)+1, then for m=1 you have 1+(k+1)=(1+k)+1. He is not assuming k+1=1+k (what he is trying to prove). – Lucas Basquerotto Apr 13 '18 at 19:04
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    @user288453: you are partly right, in saying that I assume:$ k+1 = m = 1 + k$. I took that knowledge from the comment of Eleven-Eleven, that's what I meant by confirming that I use the definition that he assumes. Of course I could have incorporated that into my answer to have more clear or complete. – C. Moos Jul 14 '18 at 12:29
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You are simply missing the last step:

\begin{align*} m+(k+1)&=(m+k)+1 \\ &=(k+m)+1\\ &=k+(1+m)\\ &=(k+1)+m \end{align*}

So you proved your inductive step.

dadexix86
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    Hi dadexix86, thank you for your reply. While scratching my head about your answer, I noticed a serious typo in my question, more specific: in the last of the manipulations. It was late here last night. I corrected it and appologize. What I am missing to bring the proof to an end, is the property $m+1=1+m$. If this was valid, the proof would finish exactly as you proposed. Unfortunately I cannot find a proof for this base step of my induction. – C. Moos May 08 '14 at 19:15