Why do you reason about the minimum number of nodes of an AVL tree of height h to argue the height is $\log n$ of an AVL tree? Recall the standard argument for showing an AVL free is of size $\log n$:


  Let $n_h = $ be the minimum number of nodes of an AVL tree of height
  $h$. Then we have:
  
  $$ n_{h} \geq 1 + n_{h-1} + n_{h-2} \implies n_h &gt; 2n_{h-2}$$ $$ n_h &gt; 2^{\frac{h}{2} } \implies h &lt; 2 \log n_h $$
  
  so the height of the tree is $O( \log n)$.


I understand the recurrence but I just don't understand why we argue about:


  be the minimum number of nodes of an AVL tree of height $h$


maybe my intuition is wrong but I thought that we'd want to argue about as many nodes as we can fit in a tree of height $h$ and show that it still balanced and show its $\log n$. Why is that reasoning incorrect?
First of all AVL trees are binary trees so you can't argue as many nodes as we can fit in a tree of height h because the maximum nodes you can fit in a AVL tree with h is 2h+2h-1+...+20. Also the minimum number of nodes on an AVL tree with h is 1+2h-1+2h-2+..+20. 

be careful for the first 1 at minimum number rather than 2h. If that plus one node would not exists, the AVL tree would be simply of height h-1, and we don't want that. 

So you have to define n as be the minimum number of nodes of an AVL tree of height h. Because your algorithm has at least that much of nodes to traverse , and you do the analyze with that assumption