If $L$ and $L^{\prime}$ are disjoint Lagrangian branes which both define the zero object in $D^{b}Fuk(X)$, then $\alpha,\beta$ are both zero, and $\mathcal{C}$ comes entirely from the $\gamma,\gamma^{\prime}$ contributions. Of course, zero Lagrangian objects have zero homology class, which cannot happen in the almost calibrated case.