or Bob, depending on the reconciliation. In direct reconciliation, we assume $P$ to be accessible to Bob, which means extending his system $B$ to $\mathbf{B}=BP$ in Eq. ([DWdr]{}). Using the equalities $S(A\mathbf{B})=S(E)$ and $S(\mathbf{B}%
|X)=S(E|X)$, we get $$K^{\ast}(\blacktriangleright)=I_{c}(A\rangle\mathbf{B})=I_{c}(A\rangle
B)+I(A,P|B),$$ where $I(A,P|B)\geq0$ is the conditional quantum mutual information. Then, in reverse reconciliation, we assume $P$ to be accessible to Alice, so that $%
A$ becomes $\mathbf{A}=AP$ in Eq. (\[DWrevEQ\]). Using $S(\mathbf{A}%
B)=S(E) $ and $S(\mathbf{A}|Y)=S(E|Y)$, we get $K^{\ast}(\blacktriangleleft
)=I_{c}(B\rangle A)+I(B,P|A)$.

Thus, the optimal key rates satisfy the inequalities$$\begin{aligned}
I_{c}(A\rangle B)~ & \leq~K(\blacktriangleright)~\leq~I_{c}(A\rangle
B)+I(A,P|B),  \label{eq1} \\
I_{c}(B\rangle A)~ & \leq~K(\blacktriangleleft)~\leq~I_{c}(B\rangle
A)+I(B,P|A),   \label{eq2}\end{aligned}$$ where the right hand sides can be bounded using$$I(A,P|B),I(B,P|A)\leq I(AB,P)\leq2\min\{S(P),S(AB)\}.$$ According to Eqs. (\[eq1\]) and (\[eq2\]), key distribution can occur ($%
K\geq0$) even in the absence of distillable entanglement ($I_{c}=0$). It is now important to note the following facts:

(i) Device-dependent QKD is the only scenario where this is possible. In fact, only in the presence of trusted noise, i.e., $S(P)>0$, we can have $%
K\geq I_{c}$ in the previous equations. Therefore device-dependent QKD is the only scenario where security may be achieved in the absence of distillable entanglement and, more strongly, in the complete absence of entanglement.

(ii) In 