[proofplan] We apply the [Signed Asymmetric Plünnecke Inequality](/theorems/???) to the pair $(A,A)$. The small-doubling hypothesis $|A+A| \leq K|A|$ gives a non-empty subset $X \subset A$ whose signed sumsets $X+mA-nA$ grow by at most the factor $K^{m+n}$. Since $X$ is non-empty, translating $mA-nA$ by any element of $X$ embeds it into $X+mA-nA$, and this transfers the estimate back to the original set $A$. [/proofplan] [step:Apply signed Plünnecke growth to obtain a good subset of $A$] For finite subsets $Y,C \subset G$ and integers $m,n \geq 0$, define \begin{align*} Y+mC-nC := \{y+c_1+\cdots+c_m-d_1-\cdots-d_n : y \in Y,\ c_i \in C,\ d_j \in C\}. \end{align*} When $m=0$ or $n=0$, the corresponding empty sum is interpreted as $0_G$, the identity element of $G$. We use the [Signed Asymmetric Plünnecke Inequality](/theorems/???): if $B,C$ are finite non-empty subsets of an abelian group and $|B+C| \leq L|B|$, then for every pair of integers $m,n \geq 0$ there exists a non-empty subset $X \subset B$ such that \begin{align*} |X+mC-nC| \leq L^{m+n}|X|. \end{align*} Apply this theorem with $B=A$, $C=A$, and $L=K$. The sets $B$ and $C$ are finite and non-empty because $A$ is finite and non-empty, and the hypothesis $|A+A| \leq K|A|$ is exactly the required growth assumption $|B+C| \leq L|B|$. Hence there exists a non-empty subset $X \subset A$ such that \begin{align*} |X+mA-nA| \leq K^{m+n}|X|. \end{align*} [guided] For finite subsets $Y,C \subset G$ and integers $m,n \geq 0$, the notation \begin{align*} Y+mC-nC \end{align*} means the set of all elements $y+c_1+\cdots+c_m-d_1-\cdots-d_n$ with $y \in Y$, each $c_i \in C$, and each $d_j \in C$. If $m=0$ or $n=0$, the corresponding sum is the empty sum $0_G$. We now apply the [Signed Asymmetric Plünnecke Inequality](/theorems/???). Its hypotheses require finite non-empty subsets $B,C$ of an abelian group and a constant $L$ such that $|B+C| \leq L|B|$. We take $B=A$, $C=A$, and $L=K$. The finiteness and non-emptiness requirements hold by the theorem statement, and the growth requirement is precisely the assumed inequality $|A+A| \leq K|A|$. Therefore, for the fixed integers $m,n \geq 0$ under consideration, the theorem supplies a non-empty subset $X \subset A$ such that \begin{align*} |X+mA-nA| \leq K^{m+n}|X|. \end{align*} The non-emptiness of $X$ is essential for the next step, because we will choose one element of $X$ and use translation by that element to compare $mA-nA$ with $X+mA-nA$. [/guided] [/step] [step:Translate $mA-nA$ into the controlled signed sumset] Choose an element $x_0 \in X$, which exists because $X$ is non-empty. Define the translation map \begin{align*} T_{x_0}: mA-nA &\to x_0+mA-nA \\ z &\mapsto x_0+z. \end{align*} Translation by $x_0$ is a bijection, so \begin{align*} |mA-nA| = |x_0+mA-nA|. \end{align*} Since $x_0 \in X$, every element of $x_0+mA-nA$ lies in $X+mA-nA$. Therefore \begin{align*} |mA-nA| = |x_0+mA-nA| \leq |X+mA-nA|. \end{align*} [guided] The Plünnecke estimate controls $X+mA-nA$, but the theorem asks for a bound on $mA-nA$. To connect the two sets, we use one element of $X$ as an anchor. Pick $x_0 \in X$. Translation by $x_0$ is a bijection on the abelian group $G$, so it preserves cardinality. Therefore \begin{align*} |mA-nA| = |x_0+mA-nA|. \end{align*} Because $x_0$ is an element of $X$, every element of the translated set $x_0+mA-nA$ is, by definition, an element of $X+mA-nA$. Thus \begin{align*} x_0+mA-nA \subset X+mA-nA. \end{align*} Taking cardinalities gives \begin{align*} |mA-nA| = |x_0+mA-nA| \leq |X+mA-nA|. \end{align*} [/guided] [/step] [step:Use $X \subset A$ to recover the stated bound in terms of $|A|$] Combining the previous step with the signed Plünnecke estimate gives \begin{align*} |mA-nA| \leq |X+mA-nA| \leq K^{m+n}|X|. \end{align*} Since $X \subset A$ and both sets are finite, \begin{align*} |X| \leq |A|. \end{align*} Hence \begin{align*} |mA-nA| \leq K^{m+n}|A|. \end{align*} This is the desired inequality for the fixed integers $m,n \geq 0$, and since $m,n$ were arbitrary, the theorem follows. [/step]