Let \(m\), \(i\) and \(n\) be positive integers, and let \((m, n) = 1\). We have the following integers:
$$ \{i, m+i, 2m+i, \ldots, (n-1)m+i \} $$
Let's say there are two positive integers \(k_1\) and \(k_1\) such that \(k_1 k_2 \lt n\) and \(k_1 m +i ≡ k_2m + i \bmod n\). This means:
$$\begin{gather} k_1 m +i ≡ k_2m + i \bmod n \\ \implies k_1 m ≡ k_2m \bmod n \end{gather}$$
According to this lemma:
$$ k_1 ≡ k_2 \bmod \frac{n}{(m,n)} $$
Since \((m,n)=1\):
$$ k_1 ≡ k_2 \bmod n $$
Since \(k_1\) and \(k_2\) and less than \(n\), then \(k_1 = k_2\). This means if two integers from \( \{ i, m+i, 2m+i, \ldots, (n-1)m+i \} \) would be conguent (modulo \(n\)), then they are the same integer. Therefore, \( \{ i, m+i, 2m+i, ..., (n-1)m+i \} \) forms a complete residue system modulo \(n\).