Let $e \in M, F: N a t \times M \rightarrow M$. Let $g: N a t \rightarrow M$ and $h: N a t \longrightarrow M$, and suppose that $g(0)=h(0)=e$. Also, suppose that, for any $k \in N a t, g(k+1)=F(k, g(k))$ and that $h(k+1)=F(k, h(k))$. Then $g=h$. [Remark. Do not just use Theorem 3-18; give your own direct proof of uniqueness using induction.]