hence the system XY is untouched. Using primes for denoting the final states we have S ( ρ$_{A}$ ) = S ( ρ$_{XY}$ ) = S ( ρ $^{′}$$_{XY}$) and S ( ρ$_{AX}$ ) = S ( ρ$_{Y}$ ) = S ( ρ $^{′}$$_{Y}$). Then we invoke the following alternative form of the strong subadditivity of the von Neumann entropy [21] to arrive at The last inequality clearly shows that the lower bound given by Eq. (17) is always smaller than the bound in (14). Hence, the inclusion of additional systems cannot possibly make an improvement in the entropic bounds of LEP. Even though Theorem 1 places some restrictions on the possible HTOs, it is possible that these may not be complete. In other words, there might be operators Q which satisfy the inequality (14) for all possible initial states ρ$_{A}$ , but it still may not be a possible HTO. Some examples of such operators will be discussed in Sec. V. However, it appears that the inequalities (14) are sufficient when E is a complete erasure operation. This subsection contains the proof of this statement. But, before that, a concise form of the inequalities in (14) should be found and for this reason it is necessary to introduce a real-valued function of hermitian operators. The needed function is the Legendre transform J of the von Neumann entropy function S . For any hermitian G , this function is defined by In order to arrive at the main property of this function, first note that for any hermitian operator G , the operator is a density matrix. Then, using the nonnegativity of S ( ρ || σ$_{G}$ ) we find 12 J ( G ) = − ln ( tr e − $^{G}$) . (20) σ$_{G}$ = exp( J ( G )) e − G (21) J ( G ) + S ( ρ ) ≤ tr ρG (22) B. Restrictions for Complete Erasure Operations S ( ρ$_{A}$ ) − S ( ρ $^{′}$$_{A}$) ≥ S ( ρ$_{AX}$ ) − S ( ρ $^{′}$$_{AX}$) . (19) S ( ρ $^{′}$$_{XY}$) + S ( ρ $^{′}$$_{XA}$) ≥ S ( ρ $^{′}$$_{Y}$) + S ( ρ $^{′}$$_{A}$) (18)