Given that $\alpha * 0 =0 $ and $\alpha * S(\beta) = \alpha *\beta + \alpha$
And we know that addition is commutative and associative.
My proof : Let $n,m \in \omega$ and i did double induction, but reached this point and could not solve.
I proved that $m*0 = 0*m = 0$
And assumed that $n*m=m*n$ and now i need to prove that $S(n)*m = m *S(n)$
Where $S(n) = n+1$
Any help is appreciated
(Getting another question off the unanswered list.)
Let $n\in\omega$; you’ve already shown that $S(n)*0=0=n*0+0$. Now suppose that $m\in\omega$ is such that $S(n)*m=n*m+m$. Then
$$\begin{align*} S(n)*S(m)&=S(n)*m+S(n)\\ &=n*m+m+S(n)\\ &=n*m+m+n+1\\ &=(n*m+n)+(m+1)\\ &=n*S(m)+S(m)\,, \end{align*}$$
where I’ve freely used commutativity and associativity of addition. It follows by induction that $S(n)*m=n*m+m$ for all $n,m\in\omega$.
Now you have the tool that you need to show that if $n,m\in\omega$ are such that $n*m=m*n$, then $S(n)*m=m*S(n)$:
$$\begin{align*} S(n)*m&=n*m+m\\ &=m*n+m\\ &=m*S(n)\,. \end{align*}$$
Since you already knew that $n*m=m*n$ when $n=0$, you can conclude by induction that $n*m=m*n$ for all $n\in\omega$.