Clear from definition.

Apply Lemma 1.4 on the definition.

Apply Lemma 1.5.

For each \(x\), replace \(\log \frac{\mathbf{P}(X_s=x)}{\mathbf{P}(Y_s=x)}\) in the definition with \(\log \frac{w_s\mathbf{P}(X_s=x)}{w_s\mathbf{P}(Y_s=x)}\) for each \(s\), and apply Lemma 1.4.

Clear from definition.

For each \(z\), \(D_{KL}(X+z\Vert Y+z)=D_{KL}(X\Vert Y)\) by Lemma 13.6. Then apply Lemma 13.5 with \(w_z=\mathbf{P}(Z=z)\).

Compare the terms correspond to each \(x\in G\) on both sides.

Clear from Definition 13.8 and Lemma 13.3.

Clear from Lemma 13.10.

For every \(G\)-valued random variable \(T\) that is independent of \(Y\),

by Lemma 1.4. Then observe that

This proves \(\ge \).

To get the equality, let \(t^*:=\arg \max _t |A \cap (H+t)|\) and observe that

Straightforward by definition and Lemma 13.14.

\(\rho ^-(U_A)\le 0\) by the choice \(T=0\). The claim then follows from Lemma 13.13.

The first claim is a direct corollary of Lemma 13.14 and Corollary 13.15. To see the second claim, observe that Lemma 13.13 and Corollary 13.15 imply \(\rho ^-(U_H),\rho ^+(U_H)\ge 0\). Therefore

which implies the second claim.

Observe that by Lemma 13.6,

Clear from definition.

The first inequality follows from Lemma 13.7. The second and third inequalities are direct corollaries of the first.

Direct corollary of Lemma 13.19.

Clear from the definition.

The first inequality follows from Lemma 13.9. The second and third inequalities are direct corollaries of the first.

Apply Lemma 13.21 for \((X,Y)\) and \((Y,X)\) and take their average.

First apply Lemma 13.25 to get \(\rho (X|X+Y)\le \rho (X) + \frac{1}{2}(\mathbb {H}[X+Y]-\mathbb {H}[Y])\), and \(\rho (Y|X+Y)\le \rho (Y)+\frac{1}{2}(\mathbb {H}[X+Y]-\mathbb {H}[X])\). Then apply Lemma 13.19 to get \(\rho (Y|X+Y)=\rho (X|X+Y)\) and take the average of the two inequalities.

Clear from compactness.

Similar to Lemma 6.12: get upper bounds for \(d[X_1;X_2]\) by \(\phi [X_1;X_2]\le \phi [X_1+X_2;\tilde X_1+\tilde X_2]\) and \(\phi [X_1;X_2]\le \phi [X_1|X_1+X_2;\tilde X_2|\tilde X_1+\tilde X_2]\), and then apply Lemma 6.8 to get an upper bound for \(I_1\).

Compare Lemma 6.8 with the identity obtained from applying Corollary 5.3 on \((X_1,\tilde X_1, X_2, \tilde X_2)\).

First of all, by \(\phi [X_1;X_2]\le \phi [X_1+\tilde X_1;X_2+\tilde X_2]\), \(\phi [X_1;X_2]\le \phi [X_1|X_1+\tilde X_1;X_2|X_2+\tilde X_2]\), and the fibring identity obtained by applying Corollary 5.3 on \((X_1,X_2,\tilde X_1,\tilde X_2)\), we have \(I_2\le \eta (d[X_1;X_2]+d[X_2;X_2])\). Then apply Lemma 13.31 to get \(I_2\le 2\eta d[X_1;X_2] +\eta (I_2-I_1)\), and rearrange.

Conditioned on every \(T_3=t\), \(d[X_1;X_2]\le d[T_1|T_3=t;T_2|T_3=t]+\eta (\rho (T_1|T_3=t)+\rho (T_2|T_3=t)-\rho (X_1)-\rho (X_2))\) by Definition 13.28. Then take the weighted average with weight \(\mathbf{P}(T_3=t)\) and then apply Lemma 3.23 to bound the RHS.

Take the average of Lemma 13.33 over all \(6\) permutations of \(T_1,T_2,T_3\).

Let \(T_1':=Y_3+Y_4\), \(T_2':=Y_2+Y_4\). First note that

On the other hand, observe that

By replacing \((Y_1,Y_2,Y_3,Y_4)\) with \((Y_1,Y_3,Y_2,Y_4)\) in the above inequalities, one has

and

Finally, take the sum of all four inequalities, apply Corollary 5.3 on \((Y_1,Y_2,Y_3,Y_4)\) and \((Y_1,Y_3,Y_2,Y_4)\) to rewrite the sum of last terms in the four inequalities, and divide the result by \(2\).

Apply Lemma 13.35 on \((Y_i,Y_j,Y_k,Y_4)\) for \((i,j,k)=(1,2,3),(2,3,1),(1,3,2)\), and take the sum.

Consider \(T_1:=X_1+X_2,T_2:=X_1+\tilde X_1, T_3:=\tilde X_1 + X_2\), and \(S=X_1+X_2+\tilde X_1 + \tilde X_2\). Note that \(T_1+T_2+T_3=0\). First apply Lemma 13.34 on \((T_1,T_2,T_3)\) when conditioned on \(S\) to get

Then apply Lemma 13.36 on \((X_1,X_2,\tilde X_1,\tilde X_2)\) and get

by Lemma 13.31. Plug in the inequality above to (1), we get

By Lemma 13.32 we can conclude that

Finally by Lemma 13.30 and \(\eta {\lt}1\) we get that the second term is \(\le 0\), and thus \(d[X_1;X_2] \le 8\eta d[X_1;X_2]\). By the choice \(\eta {\lt}1/8\) and the non-negativity of \(d\) we have \(d[X_1;X_2]=0\).

Let \(X_1,X_2\) be a \(\phi \)-minimizer. By Proposition 13.37 \(d[X_1;X_2]=0\), which by Definition 13.28 implies \(\rho (X_1)+\rho (X_2)\le \rho (Y_1) + \rho (Y_2) + \frac{1}{\eta } d[Y_1;Y_2]\) for every \(\eta {\lt}1/8\). Take the limit at \(\eta =1/8\) to get \(\rho (X_1)+\rho (X_2)\le \rho (Y_1) + \rho (Y_2) + 8 d[Y_1;Y_2]\). By Lemma 3.18 and Lemma 3.15 we have \(d[X_1;X_1]=d[X_2;X_2]=0\), and by Lemma 4.4 there are \(H_1:=\mathrm{Sym}[X_1],H_2:=\mathrm{Sym}[X_2]\) such that \(X_1=U_{H_1}+x_1\) and \(X_2=U_{H_2}+x_2\) for some \(x_2\). By Lemma 13.19 we get \(\rho (U_{H_1})+\rho (U_{H_2})\le \rho (Y_1) + \rho (Y_2) + 8 d[Y_1;Y_2]\), and thus the claim holds for \(H=H_1\) or \(H=H_2\).

Apply Proposition 13.38 on \(U_A,U_A\) to get a subspace such that \(2\rho (U_H)\le 2\rho (U_A)+8d[U_A;U_A]\). Recall that \(d[U_A;U_A]\le \log K\) as proved in Lemma 7.2, and \(\rho (U_A)=0\) by Lemma 13.17. Therefore \(\rho (U_H)\le 4\log (K)\). The claim then follows from Lemma 13.18.

Apply Corollary 13.39 and Lemma 7.1 to get a variant of Lemma 7.2. The remaining part is the same as Theorem 7.3.