Understanding the proof that A* search is optimal I don't understand the proof that $A^*$ is optimal. 

The proof is by contradiction:


  Assume $A^*$ returns $p$ but there exists a $p'$ that is cheaper. When $p$ is chosen from the frontier, assume $p''$ (Which is part of the path $p'$) is chosen from the frontier. Since $p$ was chosen before $p''$, then we have $\text{cost}(p) + \text{heuristic}(p) \leq \text{cost}(p'') + \text{heuristic}(p'')$. $p$ ends at goal, therefore the $\text{heuristic}(p) = 0$. Therefore $\text{cost}(p) \leq \text{cost}(p'') + \text{heuristic}(p'') \leq \text{cost}(p')$ because heuristics are admissible. Therefore we have a contradiction.


I am confused: can't we also assume there's a cheaper path that's in a frontier closer to the start node than $p$? Or is part of the proof that's not possible because $A^*$ would have examined that path because it is like BFS with lowest cost search, so, if there's a cheaper path, it'll be at a further frontier?
The key phrase here is


  because heuristics are admissible


In other words, the heuristics never overestimate the path length:

$$cost(n) + heuristic(n) \le cost(\text{any path going through n})$$

And since the frontier is ordered by $\textbf{cost + heuristic}$, when a completed path $p$ is dequeued from the frontier, we know that it must necessarily be $\le$ any path going through some other frontier node $q$, because

$$cost(p) = cost(p) + heuristic(p) \le cost(q) + heuristic(q) \le cost(\text{any path going through q})$$