= Solution
The initially blank register $B$ is pure and independent, so adding it changes neither relevant <Von Neumann entropy> difference: $I(A:QB)=I(A:Q)$. Bob's measurement and recording procedure, with all outcomes retained, is a <matrix trace>-preserving <quantum channel> from $QB$ to $Q'B'$. Part (a) therefore gives $I(A':Q'B')\le I(A:QB)$. Discarding $Q'$ gives $I(A':B')\le I(A':Q'B')$. Combining the two inequalities proves
$$
\boxed{I(A':B')\le I(A:Q).}
$$
Alice's register is not changed by Bob's local operation; the primes label its place in the final joint state rather than a change in its classical symbol.
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