Strong-interference information inequalities tensorize over product code inputs

First fix $x_2^n$. Under this conditioning, $Y_1^n$ and $Y_2^n$ are product channels from $X_1^n$, with the $i$th channel determined by $x_{2,i}$. Set $A^n=Y_2^n$, $B^n=Y_1^n$, and $J_i=I(X_1^n;A^i,B_{i+1}^n\mid X_2^n=x_2^n)$ for $i=0,\dots ,n$. The chain rule [F3] gives $J_i-J_{i-1}=I(X_1^n;A_i\mid A^{i-1},B_{i+1}^n,x_2^n)-I(X_1^n;B_i\mid A^{i-1},B_{i+1}^n,x_2^n)\mathrm{.}$ Because the channel is memoryless across coordinates, each term reduces to the mutual information between $X_{1,i}$ and that coordinate's output conditional on the common context $(A^{i-1},B_{i+1}^n,x_2^n)$. For each value of this context, [F1] applies to the induced law of $X_{1,i}$ and fixed $x_{2,i}$, so $J_i-J_{i-1}\ge 0$. Summing over $i$ yields $I(X_1^n;Y_2^n\mathrm{\mid }{X}_2^n=x_2^n)\ge I(X_1^n;Y_1^n\mathrm{\mid }{X}_2^n=x_2^n)$. Average over $X_2^n$.

Exchange the user indices and exchange $Y_1,Y_2$. The second strong-interference inequality in [F1] gives the same telescoping argument with receiver 1 as the more informative output for user 2. Averaging over $X_1^n$ gives $I(X_2^n;Y_1^n\mathrm{\mid }{X}_1^n)\ge I(X_2^n;Y_2^n\mathrm{\mid }{X}_1^n)$.