[proofplan]
We apply the finite-field Bogolyubov-Ruzsa theorem with quasipolynomial bounds to the set $A\subseteq \mathbb F_2^n$. The hypotheses in the present theorem are exactly the small-doubling hypotheses required by that result: $A$ is nonempty and $|A+A|\le K|A|$ with $K\ge 1$. The cited theorem then supplies a subspace contained in $2A-2A$ and gives the lower bound for its cardinality, after absorbing the absolute constant from the cited theorem into the constant $C$ appearing here.
[/proofplan]
[step:Verify the small-doubling hypotheses for the finite-field Bogolyubov-Ruzsa theorem]
Let $G := \mathbb F_2^n$ be the finite abelian group under vector addition, and let $A\subseteq G$ be the nonempty subset from the statement. Define the sumset map notation by
\begin{align*}
A+A := \{a_1+a_2 : a_1,a_2\in A\},
\end{align*}
and define the iterated difference set by
\begin{align*}
2A-2A := \{a_1+a_2-a_3-a_4 : a_1,a_2,a_3,a_4\in A\}.
\end{align*}
Because $G$ has characteristic $2$, subtraction agrees with addition, but the notation $2A-2A$ is retained to match the standard Bogolyubov-Ruzsa conclusion. The theorem statement assumes $K\ge 1$ and
\begin{align*}
|A+A|\le K|A|,
\end{align*}
so $A$ satisfies the required finite-field small-doubling hypothesis.
[guided]
The first task is only to put the hypotheses into the exact language of the finite-field Bogolyubov-Ruzsa theorem. We work in the finite abelian group
\begin{align*}
G := \mathbb F_2^n,
\end{align*}
with addition inherited from the vector-space structure. The subset $A\subseteq G$ is nonempty by hypothesis. The relevant doubling set is
\begin{align*}
A+A := \{a_1+a_2 : a_1,a_2\in A\},
\end{align*}
and the set in which we want to find a subspace is
\begin{align*}
2A-2A := \{a_1+a_2-a_3-a_4 : a_1,a_2,a_3,a_4\in A\}.
\end{align*}
Since every element of $\mathbb F_2^n$ is its own additive inverse, the signs in $2A-2A$ do not change the underlying set, but this notation is standard because the same Bogolyubov-Ruzsa statement is often formulated in arbitrary abelian groups. The assumed inequality
\begin{align*}
|A+A|\le K|A|
\end{align*}
with $K\ge 1$ is precisely the small-doubling condition needed for the finite-field Bogolyubov-Ruzsa input.
[/guided]
[/step]
[step:Apply the quasipolynomial finite-field Bogolyubov-Ruzsa theorem]
By the quasipolynomial finite-field Bogolyubov-Ruzsa theorem of Sanders, applied to the nonempty set $A\subseteq \mathbb F_2^n$ satisfying $|A+A|\le K|A|$, there is an absolute constant $C_0>0$ and a subspace $V\le \mathbb F_2^n$ such that
\begin{align*}
V\subseteq 2A-2A
\end{align*}
and
\begin{align*}
|V|\ge \exp(-C_0(\log K)^4)|A|.
\end{align*}
All hypotheses of the cited result have been verified in the previous step: the ambient group is a finite [vector space](/page/Vector%20Space) over $\mathbb F_2$, the set $A$ is nonempty, and the doubling constant is at most $K$.
[guided]
We now use the deep external input. The quasipolynomial finite-field Bogolyubov-Ruzsa theorem of Sanders states that for every nonempty subset $B\subseteq \mathbb F_2^m$ with
\begin{align*}
|B+B|\le L|B|,
\end{align*}
there is a subspace $W\le \mathbb F_2^m$ satisfying
\begin{align*}
W\subseteq 2B-2B
\end{align*}
and
\begin{align*}
|W|\ge \exp(-C_0(\log L)^4)|B|,
\end{align*}
where $C_0>0$ is an absolute constant. We apply this result with $m:=n$, $B:=A$, and $L:=K$. The previous step verified each required hypothesis: $A$ is a nonempty subset of $\mathbb F_2^n$, and the assumed bound $|A+A|\le K|A|$ supplies the doubling condition. Therefore the theorem gives a subspace $V\le \mathbb F_2^n$ with
\begin{align*}
V\subseteq 2A-2A
\end{align*}
and
\begin{align*}
|V|\ge \exp(-C_0(\log K)^4)|A|.
\end{align*}
This is the only non-elementary point of the proof; all quantitative dependence is inherited from the cited Bogolyubov-Ruzsa theorem.
[/guided]
[/step]
[step:Rename the absolute constant and conclude the stated bound]
Set $C := C_0$. Since $C_0>0$ is absolute, so is $C$. The subspace $V\le \mathbb F_2^n$ obtained above satisfies $V\subseteq 2A-2A$ and
\begin{align*}
|V|\ge \exp(-C(\log K)^4)|A|,
\end{align*}
which is exactly the desired conclusion.
[/step]
