lemma makes the reason clear:
\lemma\id{ijphase}%
-The $i$-th phase of the Iterated Jarn\'\i{}k's algorithm runs in time~$\O(m)$.
+Each phase of the Iterated Jarn\'\i{}k's algorithm runs in time~$\O(m)$.
\proof
-During the phase, the heap always contains at most~$t_i$ elements, so it takes
+During the $i$-th phase, the heap always contains at most~$t_i$ elements, so it takes
time~$\O(\log t_i)=\O(m/n_i)$ to delete an~element from the heap. The trees~$R_i^j$
are edge-disjoint, so there are at most~$n_i$ \<DeleteMin>'s over the course of the phase.
Each edge is considered at most twice (once per its endpoint), so the number
the edges of~$P_e$ must have non-increasing weights, that is $w(P_e[i+1]) \le
w(P_e[i])$.
-\alg $\<FindPeaks>(u,p,T_p,P_p)$ --- process all queries in the subtree rooted
+\alg $\<FindPeaks>(u,p,T_p,P_p)$ --- process all queries located in the subtree rooted
at~$u$ entered from its parent via an~edge~$p$.
\id{findpeaks}
\proof
Implement the Koml\'os's algorithm from Theorem \ref{verify} with the data
structures developed in this section.
-According to Lemma \ref{verfh}, it runs in time $\sum_e \O(\log\vert T_e\vert + q_e)
+According to Lemma \ref{verfh}, the algorithm runs in time $\sum_e \O(\log\vert T_e\vert + q_e)
= \O(\sum_e \log\vert T_e\vert) + \O(\sum_e q_e)$. The second sum is $\O(m)$
as there are $\O(1)$ query paths per edge, the first sum is $\O(\#\hbox{comparisons})$,
which is $\O(m)$ by Theorem \ref{verify}.