By: Jyun-Jie Liao

Congratulations on the amazing result! The proof is so elegant!

By the way, I tried the calculation a bit and I think the exponent can be improved to exactly 11 (if I didn’t do anything wrong). In the endgame, the choice of X_1',X_2' is A,B conditioned on C,S , for (A,B,C) being a permutation of (U,V,W) , and I think the bound for d[X_1^0; X_1']+d[X_2^0;X_2'] can be improved a little bits by the following argument.
First of all, Lemma 5.2 shows that

$\displaystyle d[X_1^0; A_1+A_2]+d[X_2^0; A_1+A_2]\le d[X_1^0; A_1]$
$\displaystyle +d[X_2^0; A_2]+d[A_1;A_2],$

so by taking A_1=X_1|V and A_2=X_2|S-V we get that

On the other hand, by Lemma 5.1 we have

And another formula in Lemma 5.2 shows that

$\displaystyle d[X_1^0; A_1 | A_1+A_2]+d[X_2^0; A_2 | A_1+A_2]\le d[X_1^0; A_1]$
$\displaystyle +d[X_2^0; A_2]+d[A_1;A_2],$

so by taking A_1=U,A_2=S-U we get

$\displaystyle d[X_1^0; U| V,S] + d[X_2^0; U|V,S] \le d[X_1^0; U] + d[X_2^0; S-U]$
$\displaystyle +d[U;S-U]+I(U:V|S).$

Then apply similar arguments for all 6 choices of ordered pairs in {U,V,W}, and take the average of all the inequalities. The LHS is exactly the average of d[X_1^0; X_1']+d[X_2^0;X_2] . For the RHS, first observe that the terms without X_1^0 and X_2^0 is exactly k+(I_2-I_1)/6 by Corollary 4.2. For the terms involving X_1^0 and X_2^0 , it has been shown in Section 5 and Section 6 that their sum is bounded by

$\displaystyle d[X_1^0;X_1]+d[X_2^0;X_2]+k+(I_2-I_1)/6.$

(Here we use the fact that I_2-I_1=d[X_1;X_1]+d[X_2;X_2]-2k to simplify the notation a bit.)
Therefore the average of d[X_1^0; X_1']+d[X_2^0;X_2'] -d[X_1^0;X_1]-d[X_2^0;X_2] is at most 2k+(I_2-I_1)/3 . The maximum of the upperbound for $\psi[X_1';X_2']$ again occurs at $I_1=2\eta k$ and is $8\eta k$ , so taking any $\eta<1/8$ suffices.