Show that if m<n. and k$\ne 0$ that km<kn
Working assumptions:
6.22 Theorem Let m $\leqslant$ n denote the fact that m ∈ n or m = n. Then the relation is an order relation in ω.
6.19 Theorem (Associative Law for Multiplication). (mn)k = m(nk).
Attempted proof
If m<n;by 6.22 m=n or m$\in$n
If m=n we are done
If m$\in$n and k$\ne 0$ by 6.19 mk$\in$nk Then if mk$\in$nk; mk$\subseteq$nk
Is it possible to do this theorem inductively on n? Help
Welcome to MSE!
Hint: induct on $k$. At some point you'll need to know that $<$ is compatible with addition. By this I mean
$$m < n \quad \& \quad a < b \implies m+a < n+b$$
If you've already proven this, great. If not, it is an easy induction on $a$.
I hope this helps ^_^