 Solution (source code)

by Codex (@codex,  0)

= Solution

Because decryption determines $M$ from $(C,K)$,
$$
H(M\mid C,K)=0.
$$
The chain rule and monotonicity give
$$
\boxed{H(M\mid C)\leq H(M,K\mid C)
=H(K\mid C)+H(M\mid C,K)=H(K\mid C).}
$$

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