In the above argument, once we achieved $CF^{0}(L_{1}^{\prime},L)=0$, there is a different way to proceed. We reinterpret the distinguished triangle $L_{1}^{\prime}\to L\to L^{\prime\prime}\to L_{1}^{\prime}[1]$ as an isomorphism in $D^{b}Fuk(X)$ between $L$ and a twisted complex built from $L_{1}^{\prime}\cup L^{\prime\prime}$. This would give rise to a bordism current constructed from the universal family of holomorphic curves, with $\partial\mathcal{C}=L-L_{1}^{\prime}-L^{\prime\prime}$. However, $CF^{0}(L_{1}^{\prime},L)=0$ implies that no holomorphic curve contributing to $\mathcal{C}$ passes from $L_{1}^{\prime}$ to $L$ in the clockwise direction of $\partial\Sigma$. The twisted complex structure on $L_{1}^{\prime}\cup L^{\prime\prime}$ forbids the passage from $L_{1}^{\prime}$ to $L^{\prime\prime}$ in the clockwise direction of $\partial\Sigma$. Thus if the holomorphic curve has any boundary portion on $L_{1}^{\prime}$, its entire boundary would lie on $L_{1}^{\prime}$, which cannot happen in the almost calibrated setting.