We now start with $A$, $E$ and $M$ in pure states, so that the Holevo quantity for the joint system is $\chi^{(QAEM)} = \chi^{(Q)}$. We then perform the required unitary operation between $Q$ and $A$, and a unitary operation between $A$ and $E$ which perfectly correlates the states $|kj\rangle^{(A)}$ of $A$ with orthogonal states of $E$. Unitary operations do not change the Holevo quantity. Then we trace over $E$, so that we are left with the state $$|\psi\rangle \langle\psi|^{(M)} \otimes \sum_{jk} P(k,j) |k,j\rangle \langle k,j|^{(A)} \otimes \frac{A_{kj} \rho^{(Q)} A_{kj}^\dagger}{P(k,j)} .$$ After the two unitaries and the partial trace over $E$, the Holevo quantity for the remaining systems, which we will denote by $\chi'^{(QAM)}$, satisfies $\chi'^{(QAM)} \leq \chi^{(QAEM)} = \chi^{(Q)}$. We now perform one more unitary operation, this time between $M$ and $A$, so that we correlate the states of $M$, which we denote by $|j\rangle\langle j|^{(M)}$ with the second index of the states of $A$, giving $$\sum_j |j\rangle\langle j|^{(M)} \otimes \sum_{k} P(k,j) |k,j\rangle \langle k,j|^{(A)} \otimes \sigma_{kj}^{(Q)}$$ where $\sigma_{kj}^{(Q)} = A_{kj} \rho^{(Q)} A_{kj}^\dagger/P(k,j)$ is the final state resulting from knowing both outcomes $k$ and $j$, with no knowledge of the initial choice of $i$. Finally we trace out $A$, leaving 