Prove that for any $m, n \in\omega$ that (i) $m + 0 = m$, (ii) $m + n^+ = (m + n)^+$

Question

I am getting repeatedly lost trying to approach this question:

Prove that for any $m, n \in\omega$ that (i) $m + 0 = m$, (ii) $m + n^+ = (m + n)^+$

I can fairly well grasp the idea that the definition of the addition function is an application of the Recursion Theorem, but I am getting repeatedly stumped with regard to how to demonstrate it by way of a proof. Can anyone help? Even a hint would be most welcome at this stage! Thankyou

Answer 1

According to your comments, for any $m\in\omega,$ you are defining $f_m:\omega\to\omega$ recursively by $f_m(0):=m,$ $f_m\left(n^+\right):=\bigl(f_m(n)\bigr)^+.$ Then you are defining $m+n:=f_m(n)$ for any $m,n\in\omega.$

The proof then follows directly from the definitions, since $m+0=f_m(0)=m,$ and $$m+n^+=f_m(n^+)=\bigl(f_m(n)\bigr)^+=(m+n)^+.$$