lete, because it eventually rests on an additional property, referred to as strong subadditivity. Proving strong subadditivity requires no less than two theorems, of which the proofs are quite mathematically involved! Therefore, and for the purpose of these chapters, such a property shall be taken here as a given postulate. Proof of the inequality in Eq. (X16), with discussion In the following, I simplify the notation by setting S(PA) = S(A), S(PA;PB) = S(A; B), and the like for A, B = P, Q, R. By definition, S(A; B, C) refers to the mutual infor- mation between system A and composite system BC. Then I shall observe and discuss the three following properties: (i) S(P; Q) = S(P; Q, R), since prior to measurement, R is uncorrelated to PQ and, thus, does not contribute to PQ mutual information; (ii) S(P; Q, R) > S'(P; Q, R), since the measurement between P and R has no reason to increase the mutual information between P and QR; (iii) S'(P; Q, R) > S'(P; R), since discarding the reference system Q between P and R has no reason to increase the mutual information between P and R. The first property (i) is obvious. The second property (ii) can be established as fol- lows. Let S(A; B) and S'(A;B) be the mutual information before and after the mea- surement between any systems A and B (here, A plays the role of system P and B plays the role of the composite system QR). We introduce another system C in By application of the property in Eq. (21.47), we have And because Rx is a tensor operator, we have = S(px), which, finally, yields Holevo bound a pure state (e.g., |0) or |1)). We then have S(A;B) = S(A; B, C), because of prop- erty (i). A measurement on B with the information being transferred to C must not change mutual information between A and BC, therefore, S(A; B, C) = S' (A; B, C). But on discarding C, we must have S'(A;B) < S' (A; B, C), according to property (iii), which states that discarding a system cannot increase mutual information. We must now focus on this property (iii), rewriting the inequality as S(A; B) 