The sum of triangularly disjoint graded ideals is disjoint from h

Write $x\in J$ as ${\sum }_{\beta \in S}{x}_{\beta }$ with finite support. Choose $h\in \mathfrak{h}$ for which the distinct numbers $\beta (h)$ are pairwise different. Such an $h$ exists: the product of the finitely many nonzero linear polynomials $\beta -\gamma$ is nonzero over the infinite field $\mathbb{C}$, and a nonzero polynomial cannot vanish at all complex tuples (induct on the number of variables). Applying ${\prod }_{\gamma \mathrm{\ne }\beta }(\mathrm{ad}h-\gamma (h))\mathrm{/}(\beta (h)-\gamma (h))$ to $x$ extracts $x_{\beta }$ and keeps it in $J$. Thus $J$ is graded.

If $J\cap \mathfrak{h}=0$, every vector of $J$ has zero degree-zero component by step 1.1. The algebraic sum of all these ideals consists of finite sums of their vectors, so it also has zero degree-zero component. It is an ideal because bracketing distributes over a finite sum, and it contains every such ideal. This proves existence, maximality and uniqueness of $\mathfrak{r}$.

Cartan and positive generators preserve ${\mathfrak{r}}^{+}$. Bracketing a degree $\beta >0$ vector with $f_i$ gives degree $\beta -{\alpha }_i$. If this degree is zero, the result vanishes by step 2.1; if it has mixed signs it vanishes by F1; it cannot be strictly negative unless $\beta$ were zero or a forbidden fractional multiple of ${\alpha }_i$. The remaining degree is positive. Hence ${\mathfrak{r}}^{+}$ is stable under all generators and is an ideal. The sign-changing involution proves the negative assertion. The direct sum follows from F1.