 Off-shell nilpotence of the Yang-Mills BRST quartet

Using the graded Leibniz rule, $s^{2}A=D(sc)+g(Dc)\times c=0$ because $D(c\times c)=2(Dc)\times c$. The graded Jacobi identity gives $s^{2}c=g^2(c\times c)\times c/2=0$. BRST nilpotence on the antighost and Nakanishi-Lautrup field is immediate. No field equations are required.