Talk:Natural Number Addition is Commutative/Proof 2

Clearer proof
I think maybe the section should more clearly state the property P(0) and more clearly state that induction step that if P(n) is true, P(n+) is true and that therefore, P(n) is true for all natural numbers n and therefore for all natural numbers m and n, m + n = n + m. It maybe should go something like this:

By definition of addition in ω, 0 + 0 = 0 + 0. Suppose that m + 0 = 0 + m, then m+ + 0 = m+ = (0 + m)+ = 0 + m+. Therefore, for all m, m + 0 = 0 + m.
 * m + 0 = m for all m
 * m + n+ = (m + n)+ for all m and n.

For any natural number n, suppose that, for all m, m + n = n + m. Then 0 + n+ = n+ + 0. Furthermore suppose that m + n+ = n+ + m, then we have the following so m+ + n+ = (m+ + n)+ = (n + m+)+ = (n + m)++ = (m + n)++ = (m + n+)+ = (n+ + m)+ = n+ + m+ so by induction, for all m, m + n+ = n+ + m. This shows that for any n, if for all m, m + n = n + m, then for all m, m + n+ = n+ + m so by induction, for all n, for all m, m + n = n + m so natural number addition is commutative. Blackbombchu (talk) 03:35, 25 May 2015 (UTC)
 * m + n = n + m
 * m+ + n = n + +
 * m + n+ = n+ + m


 * I'm sorry, but this seems like exactly the proof that is currently written. Except that some of the steps you wrote are relegated to other proofs. I'm currently in an ongoing project of reworking the entire natural numbers area, and this result is on my list as well. So expect its presentation to be enhanced in the coming time. &mdash; Lord_Farin (talk) 11:52, 25 May 2015 (UTC)