= Solution
Let every <root> have squared length $L$. Since $r+s$ is a root, $L=\|r+s\|^2=2L+2(r,s)$, hence $(r,s)=-L/2$. In particular the roots are independent. The potential higher sums satisfy
$$
\|2r+s\|^2=\|r+2s\|^2=3L,
$$
so neither is a root. The <root-space decomposition> therefore gives $[e_r,e_{r+s}]=[e_s,e_{r+s}]=0$.
Set $A=t\operatorname{ad}e_r$, $B=u\operatorname{ad}e_s$ and $C=[A,B]=tuN_{r,s}\operatorname{ad}e_{r+s}$. Part (iv) shows $[A,C]=[B,C]=0$. To compute the sign, differentiate $F(v)=e^{-vB}Ae^{vB}$:
$$
F'(v)=e^{-vB}[A,B]e^{vB}=C,\qquad F(0)=A.
$$
Thus $e^{-B}Ae^B=A+C$. Conjugating the <matrix exponential>, and then using the commutation of $A,C$, gives
$$
e^{-B}e^{-A}e^Be^A=e^{-A-C}e^A=e^{-C}.
$$
This proves the <central-commutator exponential identity> in the required order, and consequently the <simply laced Chevalley commutator formula> is
$$
\boxed{x_s(u)^{-1}x_r(t)^{-1}x_s(u)x_r(t)=x_{r+s}(-N_{r,s}tu).}
$$
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