Yes, thank you for this nice explanation. More formally, this seems to come
down for the heuristic to be at least admissible, or even stronger, to be
monotonic or consistent. I found that out only later on.
Given the case of an overestimating heuristic, if the program aborts upon
examination of the goal, it might miss the optimal path due to the
overestimation.
However, when it runs as long as the queue Q is not empty, I think it should
find the optimal path nevertheless, although possibly reopening the closed
goal-vertex. What do you think about that?