 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 $Q^{\prime }{B}^{\prime }$. Part (a) therefore gives $I(A^{\prime }:Q^{\prime }{B}^{\prime })\le I(A:QB)$. Discarding $Q^{\prime }$ gives $I(A^{\prime }:B^{\prime })\le I(A^{\prime }:Q^{\prime }{B}^{\prime })$. Combining the two inequalities provesAlice'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.