to obtain the class vector $$\begin{aligned}
        v_{\ell+1}&=(a_{0}^{\ell}-d_{\ell}\,;a_{1}^{\ell}-d_{\ell}, a_{2}^{\ell}-d_{\ell}, a_{3}^{\ell}-d_{\ell}, a_{4}^{\ell},\ldots,a_{n}^{\ell})\\
        &=(2a_{0}^{\ell}-a_{1}^{\ell}-a_{2}^{\ell}-a_{3}^{\ell};a_{0}^{\ell}-a_{2}^{\ell}-a_{3}^{\ell}, a_{0}^{\ell}-a_{1}^{\ell}-a_{3}^{\ell}, a_{0}^{\ell}-a_{1}^{\ell}-a_{2}^{\ell},a_{4}^{\ell},\ldots,a_{n}^{\ell})\\
        &=(a_{0}^{\ell+1};a_{1}^{\ell+1},\ldots,a_{n}^{\ell+1})
      \end{aligned}$$ and go back to Step 2.

We claim that the algorithm stops after finitely many iterations. To see this, first note that the self-intersection of all the vectors $\hat{v}_{\ell}$ is constant (since every $\hat{v}_{\ell}$ is obtained from $v_{0}$ by applying an element of $D_{K}(1,n)$). Because $v_{0}\in{\mathcal{P}}_{+}$, this implies that $$\hat{v}_{\ell}\cdot \hat{v}_{\ell} =
  (a_{0}^{\ell})^{2}-\sum_{i}(a_{i}^{\ell})^{2} \geq 0,$$ so that $|a_{0}^{\ell}|\geq|a_{i}^{\ell}|$, for all $1\leq i\leq n$. Note also that since $a_{0}^{0}\geq 0$, we must have $a_{0}^{\ell}\geq 0$, for all $\ell\geq 0$. This follows from the fact that in Step (4) above, $a_{0}^{\ell+1}=2a_{0}^{\ell}-a_{1}^{\ell}-a_{2}^{\ell}-a_{3}^{\ell}<a_{0}^{\ell}$ is negative if and only if $2a_{0}^{\ell}<a_{1}^{\ell}+a_{2}^{\ell}+a_{3}^{\ell}$. But since $$0 \leq
  A \cdot A =
  (a_{0}^{\ell})^{2}-\sum_{i}(a_{i}^{\ell})^{2}\quad\text{and}\quad (a_{1}^{\ell}+a_{2}^{\ell}+a_{3}^{\ell})^{2}\leq 3\left((a_{1}^{\ell})^{2}+(a_{2}^{\ell})^{2}+(a_{3}^{\ell})^{2}\right),$$ that would imply $$(a_{1}^{\ell})^{2} + (a_{2}^{\ell})^{2} + (a_{3}^{\ell})^{2} \leq
  (a_{0}^{\ell})^{2} \leq
  \tfrac{3}{4}\left( (a_{1}^{\ell})^{2}+(a_{2}^{\ell})^{2}+(a_{3}^{\ell})^{2} \right).$$ Hence $a_{0}^{\ell+1}$ must be non-negative.

Since the algorithm stops whenever the smallest coefficient of some $\hat{v}_{\ell}$ is negative, let us suppose that, starting with some $v_{0}\in{\mathcal{P}}_{+}$, we obtain an infinite sequence of vectors $\hat{v}_{\ell}$ with non-negative coefficients $a_{0}^{\ell}\geq a_{1}^{\ell}\geq\cdots\geq a_{n}^{\ell}\geq 0$. By Corollary \[cor:FundamentalDomain\], all the vectors $\hat{v}_{\ell}$ are in the closure of $\mathcal{C}_{K}$. Note also that none of the vectors $\hat{v}_{\ell}$ is reduced, 