Finite-type Kac–Moody roots descend to simple roots

Choose a positive symmetrizer on each finite block and combine them into $D$. F1 makes $DA$ blockwise positive definite and therefore positive definite on the full real root span. For a positive root $\beta =\sum m_i{\alpha }_i$, F2 gives $0<(\beta ,\beta )={\sum }_i{m}_i{d}_i\beta (h_i)$. Thus some $i$ with $m_i>0$ has the positive integer $\beta (h_i)>0$.

If $\beta$ is not simple, F4 implies there is some $j\mathrm{\ne }i$ with $m_j>0$. F3 makes $s_i\beta =\beta -\beta (h_i){\alpha }_i$ a root; its unchanged coefficient at $j$ is positive, so F4 forces this root to be positive. Its height has strictly decreased by the positive integer $\beta (h_i)$. Induction on positive height therefore reaches a simple root. Negative roots reduce to this case by sign, and $s_i{\alpha }_i=-{\alpha }_i$ handles the final sign.

Put $C={\max }_{i}2d_i$. By step 2.1 and form invariance, every root has squared length in $\{2d_i{\}}_i$, hence at most $C$. The positive definite matrix gives a real dual basis $v_i$ to the ${\alpha }_i$ under this form by finite elimination. For $v\mathrm{\ne }0$, the nonnegative quadratic $(x-tv,x-tv)$ at $t=(x,v)\mathrm{/}(v,v)$ gives $(x,v{)}^2\le (x,x)(v,v)$. If $\beta =\sum m_i{\alpha }_i$, then $\mathrm{\mid }{m}_i{\mathrm{\mid }}^2=\mathrm{\mid }(\beta ,v_i){\mathrm{\mid }}^2\le C(v_i,v_i)$. Each integer coordinate therefore belongs to a fixed finite interval. There are only finitely many such tuples, proving $\mathrm{\mid }\mathrm{\Delta }\mathrm{\mid }<\mathrm{\infty }$.

Every root is real by step 2.1, so F5 gives dimension one. Each finite block is invertible by F1; hence $\mathrm{rank}A=n$ and the minimal Cartan has dimension $n$. Summing the root decomposition of F4 gives $\dim \mathfrak{g}=n+\mathrm{\mid }\mathrm{\Delta }\mathrm{\mid }$, finite by step 3.1.