It is clear that

$$\operatorname{div}(t)=D_{0}-D_{\infty}+\sum_{i=0}^{k}a_{i}E_{i}+\sum_{j\in\Theta_{i}}a_{i,j}F_{i,j}$$

for certain coefficients $a_{i}$ and $a_{i,j}$ that we want to determine as much as possible.

If a component $E$ of $\mathcal{X}_{0}$, with coefficient $a$, does not meet $D_{0}$ nor $D_{\infty}$, but meets $r\geq 1$ other components, and the coefficients of $r-1$ of these components are equal to $a$, while the coefficient of the remaining component is $b$, we obtain that

$$0=\operatorname{div}(t)\cdot E=aE\cdot E+a(r-1)+b=-ra+a(r-1)+b=b-a$$

Thus $b=a$. Starting with the components $F_{i,j}$ that are terminal, we deduce that, for all $i$ and $j\in\Theta_{i}$, $a_{i}=a_{i,j}$. Therefore,

$$\operatorname{div}(t)=D_{0}-D_{\infty}+\sum_{i=0}^{k}a_{i}\left(E_{i}+\sum_{j\in\Theta_{i}}F_{i,j}\right).$$

In particular, the lemma is proved for $k=0$. Assume now that $k>0$.

It only remains to show that $a_{i}=a_{0}-i$, that we prove by induction. For $i=1$, we compute

$$0=\operatorname{div}(t)\cdot E_{0}=D_{0}\cdot E_{0}+a_{0}E_{0}\cdot E_{0}+a_{0}\sum_{j\in\Theta_{0}}F_{0,j}\cdot E_{0}+a_{1}E_{1}\cdot E_{0}=1-a_{0}+a_{1}.$$

Thus $a_{1}=a_{0}-1$. For $1<i\leq k$, by induction hypothesis, $a_{i-1}=a_{i-2}-1$. Then

$$0=\operatorname{div}(t)\cdot E_{i-1}=a_{i-2}-2a_{i-1}+a_{i}=1-a_{i-1}+a_{i}.$$

Thus $a_{i}=a_{i-1}-1=a_{0}-i$, proving the lemma. ∎