Replacing Bob's key by a copy of Alice's costs the correctness error

The blocks of ${\rho }_{ATE}$ are the finite sums ${\sum }_b{R}_E^{a,b,t}$, so they are nonnegative. Their traces sum to the trace of ${\rho }_{ABTE}$ by [F1], hence to one; therefore ${\rho }_{ATE}$ is a density operator. Complete positivity and trace preservation [F2] show that ${\gamma }_{ABTE}=\mathcal{C}({\rho }_{ATE})$ is also a density operator, as required for [F3].

By [F2], ${\gamma }_{ABTE}$ has block ${\sum }_b{R}_E^{a,b,t}$ in each key-diagonal slot $(a,a,t)$ and zero in every slot $(a,b,t)$ with $b\mathrm{\ne }a$. Thus ${\rho }_{ABTE}-{\gamma }_{ABTE}$ has block $R_E^{a,b,t}$ in each slot with $a\mathrm{\ne }b$, block $-{\sum }_{b\mathrm{\ne }a}{R}_E^{a,b,t}$ in slot $(a,a,t)$, and zero elsewhere. Each displayed nonzero block is self-adjoint because it is a positive block or its negative.

Apply [F4] to these orthogonal classical blocks. For $a\mathrm{\ne }b$, [F5] gives $\mathrm{\parallel }{R}_E^{a,b,t}{\mathrm{\parallel }}_1=\mathrm{tr}(R_E^{a,b,t})$. The sum ${\sum }_{b\mathrm{\ne }a}{R}_E^{a,b,t}$ is nonnegative, so [F5] and [F6] give $\mathrm{\parallel }-{\sum }_{b\mathrm{\ne }a}{R}_E^{a,b,t}{\mathrm{\parallel }}_1={\sum }_{b\mathrm{\ne }a}\mathrm{tr}(R_E^{a,b,t})$. Therefore $\mathrm{\parallel }{\rho }_{ABTE}-{\gamma }_{ABTE}{\mathrm{\parallel }}_1=2{\sum }_{a\mathrm{\ne }b,t}\mathrm{tr}(R_E^{a,b,t})$. Dividing by $2$ as in [F3] proves the equality with the mismatch probability.