%! TEX root = EMC.tex
% vim: tw=50
% 13/03/2025 09AM
%
So \cref{coro:7.8} now gives us:
\begin{align*}
  \entd{X_1}{X_3} + \entd{X_2}{X_4}
  &\ge \entd{X_1 + X_2}{X_3 + X_4}
  + \entd{X_1 \mid X_1 + X_2}{X_3 \mid X_4} \\
  &~~\third(\scentd{X_1 + X_2}{X_1 + X_3}{X_2 + X_3,
  W} \\
  &~~+ \cent{X_1 + X_2}{W} + \cent{X_1 + X_3}{W}
  - \cent{X_2 + X_3}{W}
  )
\end{align*}
Now apply this to $(X_1, X_2, X_3, X_4)$, $(X_1,
X_2, X_4, X_3)$ and $(X_1, X_4, X_3, X_2)$ and
add.

We look first at the entropy terms. We get
\begin{align*}
  &2\cent{X_1 + X_2}{W}
  + \cent{X_1 + X_4}{W}
  + \cent{X_1 + X_3}{W}
  + \cent{X_1 + X_4}{W}
  + \cent{X_1 + X_2}{W} \\
  &~~- 2\cent{X_1 + X_2}{W}
  - 2\cent{X_2 + X_4}{W}
  - 2\cent{X_1 + X_2}{W} \\
  &= 0
\end{align*}
where we made heavy use of the observation that if
$i, j, k, l$ are some permutation of $1, 2, 3, 4$,
then
\[
  \cent{X_i + X_j}{W}
  = \cent{X_k + X_l}{W}
.\]
This also allowed use e.g. to replace
\[
  \scentd{X_1 + X_2, X_3 + X_4}{X_1 + X_3, X_2 +
  X_4}{X_2 + X_3, W}
\]
by
\[
  \scentd{X_1 + X_2}{X_1 + X_3}{X_2 + X_3, W}
.\]
Therefore, we get the following inequality:

\begin{fclemma}[]
  \label{lemma:7.9}
  % Lemma 7.7
  \begin{align*}
    &\mathcloze[1]{2\entd{X_1}{X_2}
    + 2\entd{X_3}{X_4}
    + \entd{X_1}{X_4}
    + \entd{X_2}{X_3}} \\
    &\ge \mathcloze[2]{2\entd{X_1 + X_2}{X_3 + X_4}
    + \entd{X_1 + X_4}{X_2 + X_3}} \\
    &~~+ \mathcloze[3]{2\centd{X_1}{X_1 + X_2}{X_3}{X_3 + X_4}
    + \centd{X_1}{X_1 + X_4}{X_2}{X_2 + X_3}} \\
    &~~+ \mathcloze[4]{\third\bigg(
      \scentd{X_1 + X_2}{X_1 + X_3}{X_2 +
      X_3, W}
      + \scentd{X_1 + X_2}{X_1 + X_4}{X_2 +
      X_4, W}} \\
    &~~+ \mathcloze[5]{\scentd{X_1 + X_4}{X_1 + X_3}{X_3 + X_4, W}
    \bigg)}
  \end{align*}
\end{fclemma}

\begin{proof}
  Above.
\end{proof}

Now let $X_1, X_2$ be copies of $X$ and $Y_1, Y_2$
copies of $Y$ and apply \cref{lemma:7.9} to $(X_1,
X_2, Y_1, Y_2)$ (all independent), to get this.

\begin{fclemma}[]
  \label{lemma:7.10}
  % Lemma 7.8
  Assuming:
  - $X_1, X_2, Y_1, Y_2$ satisfy: $X_1$ and $X_2$
  are copies of $X$, $Y_1$ and $Y_2$ are copies of
  $Y$, and all of them are independent
  Then:
  \begin{align*}
    &\mathcloze{6\entd XY} \\
    &\mathcloze{\ge 2\entd{X_1 + X_2}{Y_1 + Y_2}
    + \entd{X_1 + Y_2}{X_2 + Y_1}} \\
    &~~\mathcloze{+ 2\centd{X_1}{X_1 + X_2}{Y_1}{Y_1 + Y_2}
    + \centd{X_1}{X_1 + Y_1}{X_2}{X_2 + Y_2}} \\
    &~~\mathcloze{+ \frac{2}{3} \scentd{X_1 + X_2}{X_1 +
    Y_1}{X_2 + Y_1, X_1 + Y_2}} \\
    &~~\mathcloze{+ \third\scentd{X_1 + Y_1}{X_1 + Y_2}{X_1 + X_2,
    Y_1 + Y_2}}
  \end{align*}
  OR? TODO: figure out which is correct
  \begin{align*}
    &6\entd XY \\
    &\ge 2\entd{X_1 + X_2}{Y_1 + Y_2}
    + \entd{X_1 + Y_1}{X_2 + Y_2} \\
    &~~+ 2\centd{X_1}{X_1 + X_2}{Y_1}{Y_1 + Y_2}
    + \centd{X_1}{X_1 + Y_1}{X_2}{X_2 + Y_2} \\
    &~~+ \frac{2}{3} \centd{X_1 + X_2}{X_1 + Y_1}{X_2
    + Y_1}{X_1 + Y_2} \\
    &~~+ \frac{1}{3} \centd{X_1 + Y_1}{X_1 + Y_2}{X-1
    + X_1}{Y_1 + Y_2}
  \end{align*}
\end{fclemma}

\begin{proof}
  Use above.
\end{proof}

Recall that we want $(U, V)$ such that
\begin{align*}
  \tau_{X, Y}[U; V]
  &= \entd UV
  + \eta(\entd UX + \entd VY) \\
  &< \entd XY
\end{align*}
\cref{lemma:7.10} gives us a collection of
distances (some conditioned), at least one of
which is at most $\frac{6}{7} \entd XY$. So  it
will be enough to show that for all of them we get
\[
  \entd UX
  + \entd VY
  \le C \entd XY
,\]
for some absolute constant $C$. Then we can take
$\eta < \frac{1}{7C}$.

\begin{fcdefnstar}[$C$-relevant]
  \glsadjdefn{Crel}{$C$-relevant}{variables}%
  Say that $(U, V)$ is \emph{$C$-relevant} to $(X,
  Y)$ if
  \[
    \entd UX
    + \entd VY
    \le C \entd XY
  .\]
\end{fcdefnstar}

\begin{fclemma}[]
  \label{lemma:7.11}
  % Lemma 7.9
  $(Y, X)$ is \cloze{\glsref[Crel]{$2$-relevant} to $(X,
  Y)$.}
\end{fclemma}

\begin{proof}
  $\entd YX + \entd XY = 2\entd XY$.
\end{proof}

\begin{fclemma}[]
  \label{lemma:7.12}
  % Lemma 7.10
  Assuming:
  - $U, V, X$ be independent $\Fbb_2^n$-valued random
  variables
  Then:
  \[
    \entd{U + V}{X}
    \le \half (
      \entd UX
      + \entd VX
      + \entd UV
    )
  .\]
\end{fclemma}

\begin{proof}
  \begin{align*}
    \entd{U + V}{X}
    &= \ent{U + V + X}
    - \half \ent{U + V}
    - \half \ent X \\
    &= \ent{U + V + X}
    - \ent{U + V}
    + \half \ent{U + V}
    - \half \ent X \\
    &\le \half \ent{U + X}
    - \half \ent U
    + \half \ent{V + X}
    - \half \ent V
    + \half \ent{U + V}
    - \half \ent X \\
    &= \half(
      \entd UX
      + \entd VX
      + \entd UV
    )
    \qedhere
  \end{align*}
\end{proof}

\begin{fccoro}[]
  \label{coro:7.13}
  % Corollary 7.11
  Assuming:
  - $(U, V)$ is \gls{Crel} to $(X, Y)$
  - $U_1, U_2, V_1, V_2$ are independent copies of $U, V$
  Then:
  $(U_1 + U_2, V_1 + V_2)$ is
  \glsref[Crel]{$2C$-relevant} to $(X, Y)$.
\end{fccoro}

\begin{proof}
  \begin{align*}
    &\entd{U_1 + U_2}{X}
    + \entd{V_1 + V_2}{Y} \\
    &\le \half(
      2\entd UV
      + \entd UU
      + 2\entd VY
      + \entd VV
    )
    &&\text{(by \cref{lemma:7.12})} \\
    &\le 2(
      \entd UX
      + \entd VY
    )
    &&\text{(by \nameref{lemma:6.7})} \\
    &\le 2C\entd XY
  \end{align*}
\end{proof}

\begin{fccoro}[]
  \label{coro:7.14}
  % Corollary 12
  \cloze{$(X_1 + X_2, Y_1 + Y_2)$} is
  \cloze{\glsref[Crel]{$4$-relevant} to $(Y, X)$.}
\end{fccoro}

\begin{proof}
  $(X, Y)$ is \glsref[Crel]{$2$-relevant} to
  $(Y, X)$, so by \cref{coro:7.13} we're done.
\end{proof}

\begin{fccorostar}[]
  % \label{coro:7.15}
  % Corollary 13
  Assuming:
  - $(U, V)$ is \gls{Crel} to $(X, Y)$
  Then:
  $(U + V, U + V)$ is \glsref[Crel]{$(3C +
  2)$-relevant} to $(X, Y)$.
\end{fccorostar}

\begin{proof}
  By \cref{lemma:7.12},
  \begin{align*}
    \entd{U + V}X
    + \entd{U + V}Y
    &\le \half \big(
      \entd UX
      + \entd VX
      + \entd UY
      + \entd VY
      + 2\entd UV
    \big) \\
    &\le \half (
      2\entd UX
      + 4\entd UV
      + 2\entd VY
    ) \\
    &\le \half(
      6\entd UX
      + 6\entd VY
      + 4\entd XY
    )
    \qedhere
  \end{align*}
\end{proof}