By the triangle inequality we have \(|A-B| \leq \frac{|A+B| |B+B|}{|B|}\), and from the Plünnecke-Ruzsa inequality we have \(|B+B| \leq (\frac{|A+B|}{|A|})^2 |A|\).

This is obvious from the bijection of multiplication by \(x\).

Trivial from the definitions.

For any value \((a+b) v\) with \(v \in A\) we have \(a v \in a A\), \(b v \in b A\), and \(a v + b v = (a+b)v\)

Exactly the same as the previous lemma.

By the triangle inequality, we have \(|B+C| \leq \frac{|B + (A \cap C)| |(A \cap C) + C|}{|A \cap C|}\), and this is less than \(\frac{|B+A| |C+C|}{|A \cap C|}\) because \(B + (A \cap C)\subseteq B+A\) and \((A \cap C) + C \subseteq C + C\).

If either \(A\) or \(B\) are empty this is trivial. Otherwise we have an element \(v \in B\). We obviously have \((A + B) \cap (A + \{ v\} ) = A + \{ v\} \), and it is nonempty. So by 1.6 we have

By Ruzsa’s covering lemma we have a set \(u\) of size \(\leq \frac{|A+B|}{|B|}\) such that \(A \subseteq u + B - B\). This gives \(\frac{|C + A| |(A+B) + (A+B)|}{|A|} \leq \frac{|C + A| |(u+B-B+B) + (u+B-B+B)|}{|A|} = \frac{|C + A| |2 \cdot u + 4 \cdot B - 2 \cdot B|}{|A|} \leq \frac{|C + A| |u|^2 |4 \cdot B - 2 \cdot B|}{|A|}\). By the Plünnecke-Ruzsa inequality we now have \(|4 \cdot B - 2 \cdot B| \leq (\frac{|A+B|}{|A|})^6 |A|\), and the result follows from this and the bound on \(|u|\).

By a direct bijection from quadruples \((a, b, c, d) \in A \times (xA) \times A \times (xA)\) such that \(a + b = c + d\).