Since most parts of the proof are identical to the distinguished triangle case, we will only sketch the main difference.

The Lagrangian boundaries on $\partial\Sigma$ are arranged in the clockwise order as $L,L_{N},L_{N-1},\ldots L_{1}$. We construct the holomorphic function $F$ as in (33), and use it to produce complex valued volume forms on the $(n-1)$-dimensional moduli spaces, such that for any $m=1,\ldots N$,

$$\int_{L}\Omega=\int_{\mathcal{M}}\tilde{\Omega}_{L},\quad\int_{L_{i}}\Omega=\int_{\mathcal{M}}\tilde{\Omega}_{L_{i}},\quad i=1,2,\ldots N.$$

As before, the real part of these complex volume forms are all non-negative, as a consequence of the positivity condition. The claim on the image curve $F(\Sigma)$ holds verbatim. Similarly to the distinguished triangle case, we produce nonnegatively weighted subsets $A_{1},\ldots A_{N}\subset L$ with $L=\sum A_{i}$, such that

$$\text{Re}\int_{A_{m}}\Omega=\text{Re}\int_{L_{m}}\Omega>0,\quad\text{Im}\sum_{1}^{m}\int_{A_{i}}\Omega\geq\text{Im}\sum_{1}^{m}\ \int_{L_{i}}\Omega,\quad\int_{L}\Omega=\sum_{1}^{N}\int_{L_{i}}\Omega.$$

(37)

This implies

$$\arg\int_{A_{1}}\Omega\geq\arg\int_{L_{1}}\Omega,\quad\arg\int_{A_{N}}\Omega\leq\arg\int_{L_{N}}\Omega,$$

whence the phase inequality (35).

The J-volume can be bounded below by

$$\text{Vol}_{J}(L)=\int_{L}|\Omega|=\sum_{1}^{N}\int_{A_{i}}|\Omega|\geq\sum_{1}^{N}|\int_{A_{i}}\Omega|\geq\sum_{1}^{N}|\int_{L_{i}}\Omega|.$$

The last step uses the purely numerical Lemma 3.27 below. ∎