Tree induction

Formally, assume $a\text{tree}$ means that $a$ is a binary tree. To prove that whenever $a\text{tree}$, we have $𝒫(a)$, it's sufficient to show that:

1. $𝒫(\text{empty)}$.

2. For every $a_1$ and $a_2$ such that $a_1\text{tree}$ and $a_2\text{tree}$, if $𝒫(a_1)$ and $𝒫(a_2)$ then $𝒫(\text{node}(a_1;a_2))$.