I was reading BFT from CLRS. This is the example given in the book:
(Figure:The operation of BFS on an undirected graph. Tree edges are shown shaded as they are produced by BFS. The value of $u.d$ appears within each vertex $u$. The queue $Q$ is shown at the beginning of each iteration. Vertex distances appear below vertices in the queue.)
Then it gives following lemma:
Suppose that during the execution of BFS on a graph $G=(V,E)$, the queue $Q$ contains the vertices $<v_1,v_2,...,v_r>$, where $v_1$ is the head of $Q$ and $v_r$ is the tail. Then, $v_r.d\leq v_1.d+1$ and $v_i.d\leq v_{i+1}.d$ for $I=1,2,...,r-1$.
I am ok with $v_i.d\leq v_{i+1}.d$ for $I=1,2,...,r-1$ as I can observe this in all queue instances above. But I am not able to understand how $v_r.d\leq v_1.d+1$ aligns with above, as $i<i+1$, but $r>1$. How that $+1$ is making whole difference?