Only the highest dot orbit can occur in the integrable numerator

Let $\eta =\lambda -\beta$ have a nonzero coefficient, so $\beta \in Q^{+}$ by F2. All its simple labels are integers by F3 and the integral Cartan matrix. If $\eta (h_i)=0$, reflection fixes $\eta$ and F1 makes its coefficient its negative, impossible over $\mathbb{Z}$. If $\eta (h_i)<0$, reflect: $s_i\eta =\eta +k{\alpha }_i$ for the positive integer $k=-\eta (h_i)$. Its coefficient is still nonzero by F1, so F2 implies $\beta -k{\alpha }_i\in Q^{+}$. Its height is smaller by $k$. Repeatedly choosing the least negative index terminates in finitely many steps, at a support weight ${\eta }^{\mathrm{\prime }}$ with every label strictly positive. This termination uses the actual cone bound on the orbit's support, not a claim that every integral weight can be moved to the dominant chamber.

Write ${\eta }^{\mathrm{\prime }}=\lambda -\gamma$, $\gamma ={\sum }_i{k}_i{\alpha }_i\in Q^{+}$. F2 says ${\eta }^{\mathrm{\prime }2}={\lambda }^2$. But F6 gives ${\lambda }^2-{\eta }^{\mathrm{\prime }2}=(\gamma ,\lambda +{\eta }^{\mathrm{\prime }})={\sum }_i{k}_i{d}_i(\lambda (h_i)+{\eta }^{\mathrm{\prime }}(h_i))\mathrm{.}$ Since $\lambda (h_i)=\mathrm{\Lambda }(h_i)+1\ge 1$ and ${\eta }^{\mathrm{\prime }}(h_i)>0$, this is strictly positive unless every $k_i=0$. Hence ${\eta }^{\mathrm{\prime }}=\lambda$. Step 1.1 now puts every support weight on the orbit of $\lambda$. The comparison is between real sums of labels even if complementary Cartan coordinates are complex. No positive-definiteness assumption was made.

For a reduced word $w=s_{i_1}\cdots s_{i_t}$, telescoping gives $\lambda -w\lambda ={\sum }_{j=1}^t\lambda (h_{i_j})s_{i_1}\cdots s_{i_{j-1}}{\alpha }_{i_j}\mathrm{.}$ Every prefix is reduced, and its next root is positive by F5. Every scalar is an integer at least one by F3. Thus the difference belongs to $Q^{+}$ and has height at least $t=\mathrm{\ell }(w)$, using F4's length convention. If $w\lambda =\lambda$, this forces $t=0$, so the stabilizer is trivial. A fixed height bound allows only finitely many words in the finite alphabet, proving local finiteness. F2 gives coefficient one at $\lambda$; F1 then gives exactly $\det (w)$ at $w\lambda$. Together with step 2.1 this proves all assertions. Empty words and the empty simple system give the sole term of coefficient one. No infinite choices occur.