which is a constant independent of the device state ρ . Let us first note that, by using the identity in Eq. (15), it is possible to express Δ as which clearly shows that Δ is a strictly positive number. It will be argued below that the sequence { ν$_{k}$ } can be selected in such a way that Δ can be adjusted to be equal to any positive number and this can be achieved with a finite Z$_{B}$ . This will be done by showing that Δ can be made arbitrarily large and arbitrarily small and then allude to continuity to prove the claim. (1) First, note that if the parameter ν$_{2}$ of X$_{2}$ is increased towards 1, the final energy of X$_{2}$ approaches to infinity. Hence, in this limit Δ diverges to ∞ . Therefore, Δ can be made arbitrarily large. (2) In order to make Δ small, the parameters of neighboring subsystems should be chosen nearly equal. In other words, the increments ν$_{k}$$_{+1}$ − ν$_{k}$ should tend to 0. In this limit, the series in Eq. (A7) approach to an integral, Using h (0) = Q − q$_{0}$ 1 , it can be seen that Δ approaches to 0 in this limit. This shows that Δ can be chosen to be as small as possible. Moreover, for any given parameter sequence { ν$_{k}$ } it is possible to adjust the large k behavior of these sequences and make Z$_{B}$ finite. For example, such an adjustment can be made only for indices k that exceed a threshold k$_{min}$ . By selecting the threshold k$_{min}$ sufficiently large, it is possible to insure that Δ does not change much. In summary, therefore, Δ can be made as small as possible by keeping Z$_{B}$ finite. Since Δ depends continuously on ν$_{k}$ , it is then possible to adjust these parameter se- quences such that Δ has the value Δ = k$_{B}$T J , which is a strictly positive quantity by the initial supposition. Then, by Eq. (A6), the change in the energy of the bath is Δ E$_{B}$ = tr ρQ and hence Q is the HTO associated with this realization. This completes the proof of the Theorem 3. 26 ∞ ∑ k =2 tr( σ$_{k}$$_{−}$$_{1}$ − σ$_{k}$ ) H$_{X}$$_{k}$ −→ − ∫ 1 0 dν tr ( ∂σ ( ν ) ∂ν h ( ν ) ) = − k$_{B}$T ∫ 1 0 dν ∂S ( σ ( ν )) ∂ν = k$_{B}$T S ( σ (0)) . (A9) Δ = k$_{B}$T ∞ ∑ k =1 S ( σ$_{k}$ || σ$_{k}$$_{+1}$ ) , (A8)