of ${{\mathfrak{g}}}$ generated by $\{E_i,F_i,H_i\ |\ 1\leq i\leq n\}$ is a Kac-Moody algebra with Cartan matrix $C = [C_{ij}] = [2(\beta_i,\beta_j)/(\beta_j,\beta_j)]$. Denote this subalgebra by ${{\mathfrak{g}}}(\beta_1,\cdots,\beta_n)$.

By construction, the elements $H_i$ are in the Cartan subalgebra, ${{\mathfrak{h}}}$, so the following relations are clear: $$[H_i,E_j] = \beta_j(H_i) E_j, \quad
[H_i,F_j] = -\beta_j(H_i) F_j, \quad
[H_i,H_j] = 0.$$ Because all of the $\beta_i$ are positive real roots, the matrix $C$ satisfies the conditions to be a Cartan matrix. For $i\neq j$, the bracket $[E_i,F_j]$ would be in the $\beta_i - \beta_j$ root space, but we have chosen the $\beta_i$ such that this difference is not a root of ${{\mathfrak{g}}}$, so that bracket must be zero, giving the relations $$[E_i,F_j] = \delta_{ij} H_i.$$ To check that $(ad\ E_i)^{1-C_{ji}}\ E_j = 0$ for all $i\neq j$, it would suffice to show that $(1 - C_{ji})\beta_i + \beta_j$ is not in $\Phi$. Since $\beta_j$ is a real root of ${{\mathfrak{g}}}$, and since $\beta_j - \beta_i\notin\Phi$, the $\beta_i$ root string through $\beta_j$ is of the form $$\beta_j, \beta_j + \beta_i, \cdots, \beta_j + q\beta_i$$ where $-q = 2(\beta_j,\beta_i)/(\beta_i,\beta_i) = C_{ji}$. Therefore, $$\beta_j + (q+1)\beta_i = \beta_j + (1 - C_{ji})\beta_i$$ is not in $\Phi$. The relations 