Let's consider pair of vertices $e$ and $h$. For them we can safely lay, from the BFS tree, following properties:
- lower bound: $2$
- upper bound: $4$
By having a look at the graph we started from, we can see that we have a path ‹$e, j, h$› that has a length 2. Apart from that we can also notice there is another path from $e$ to $h$ and that is ‹$e, a, c, i, d, h$›. And that path has a length of $5$. Doesn't this break our statements at the beginning? (_I'm leaving that as an exercise ;)_)
## Proof by contradiction
Let's keep the same graph, but break the lower bound, i.e. I have gotten a lower bound $2$, but „there must be a shorter path“! ;)
Now the more important question, is there a shorter path in that graph? The answer is no, there's no shorter path than the one with length $2$. So what can we do about it? We'll add an edge to have a shorter path. Now we have gotten a lower bound of $2$, which means the only shorter path we can construct has $1$ edge and that is ‹$e, h$› (no intermediary vertices). Let's do this!
