Don't understand one step for AVL tree height log n proof
You can continue as same as line 4 the process like that:

$$ N_h &gt; 2N_{h-2}&gt; 2(2 N_{h-4})&gt;2(2(2 N_{h-6}))&gt;\cdots$$
As you can see, the indexs are decreasing by substracting $2$ in each step when you use the inequality. So, the process stops when the index $h$ takes $0$, but from the indexs behavior the half of $h$ (floor) will be the quantity of times that we substract $2$ from $h$.

$$ N_h &gt; 2N_{h-2}&gt;\cdots&gt;2(2(2(2(2h^{h-10}))))&gt; 2^{\frac{h}{2}}$$