Application of Davenport theorem The Davenport theorem (or Cauchy-Davenport theorem for some authors ) states that for any two nonempty
subsets $A$ and $B$ of the prime field $\mathbb Z/p\mathbb Z$ we have $$|A+B| ≥ \min(p, |A|+|B|−1),$$ where
$A+B := \{a+b \;\; \mod p \;:\; a ∈ A, b ∈ B\}$.
In the following paper by P. Erdos and H. Heilbronn 
https://www.renyi.hu/~p_erdos/1964-18.pdf
the authors apply Davenport theorem in the proof of$\;$ LEMMA I.1. 
I can not understand how Davenport theorem is applied here. Can someone can help me?
 A: In the paper, Lemma I.1 is stated as the following.

LEMMA I.1. Let $1 < m\le l < \frac12p$; $a_1,\ldots,a_m$ are distinct non-zero  residue classes mod $p$. Then there exists an $i$ in $1 \le i \le k$ such that  $B(a_i) < l-\frac16 m$.
Proof. Put $r = 1+[2l/m]$. By Davenport’s theorem [1] about the  addition of residue classes mod $p$, applied to the residue classes $0, a_1,\ldots, a_m$,  we obtain $t\ge \min(p - 1, rm)$ distinct non-zero residue classes $c_1,\ldots,c_t$  which can be expressed as the sum of at most $r$ residue classes $a_j$ ($1 \le  j \le m$), which need not have distinct indices $j$.
$$\vdots$$

The Davenport theorem is applied $r-1$ times.
Let $A=\{0,a_1\ldots,a_m\}$. For simplicity, write $\underbrace{A+\cdots+A}_\text{n times}=nA$. Then by the Davenport theorem, \begin{align}|rA|=|A+(r-1)A|&\ge\min(p,|A|-1+|(r-1)A|)\\&=\min(p,|A|-1+|A+(r-2)A|)\\&\ge\min(p,|A|-1+\min(p,|A|-1+|(r-2)A|))\\&=\min(p,2|A|-2+|(r-2)A|)\\&\ge\cdots\\&=\min(p,r|A|-(r-1))\\&=\min(p,rm+1)\end{align} All elements of $rA$ except $0$ is expressed as the sum of at most $r$ residue classes $a_j$. So by excluding $0$, we get $rA-\{0\}=\{c_1,c_2,\ldots,c_t\}$ where $c_1,\ldots,c_t$ are distinct non-zero residue classes which can be expressed as the sum of at most $r$ residue classes $a_j$. From the inequality above, we have $t=|rA|-1\ge\min(p-1,rm)$.
