Prove correctness of recursive multiplication algorithm
You need to do a proof by induction on $n + r$. Let $P(m)$ be the statement "for any $n,r \geq 1$ satisfying $n+r = m$, $\mathrm{multiply}(n,r) = nr$. The proof by induction goes along the following lines.

Base case: $m=2$, so $n=r=1$. By inspection, $\mathrm{multiply}(1,1)=1$.

Induction step: suppose $P(m)$ is true for some $m \geq 2$, and let $n,r\geq 1$ satisfy $n+r = m+1$. If $n=1$ or $r=1$ then $\mathrm{multiply}(n,r)=nr$ by inspection. Otherwise, by $P(m)$, $\mathrm{multiply}(n-1,r)=(n-1)r$ (since $n-1+r = m$), so $\mathrm{multiply}(n,r) = r+\mathrm{multiply}(n-1,r) = r+(n-1)r = nr$.

By the principle of mathematical induction, $P(m)$ is true for all $m$, and so for any $n,r\geq 1$, $\mathrm{multiply}(n,r)=nr$.