 Solution

Factor the odd parameter on the left, $\delta =ϵs$. The resulting left-acting BRST differential obeys the graded Leibniz ruleFor the odd Grassmann field $c$, the bracket in the transformation is a graded commutator: $[c,c{]}_{\mathrm{gr}}=2c^2$, not the identically zero ordinary commutator of a matrix with itself. Thus $sc=i{c}^2$, while $s{A}_{\mu }=D_{\mu }c$, $s\bar{c}=h$ and $sh=0$.

On the ghost,On the gauge field, variation of the connection and the adjoint covariant derivative givesHere $D_{\mu }$ is even and therefore obeys the ordinary product rule. Also $s^2\bar{c}=sh=0$ and $s^{2}h=0$, without using any field equation; this is off-shell nilpotence supplied by the Nakanishi-Lautrup field.

Applying the graded Leibniz rule twice cancels the two cross terms:The square is consequently an even graded derivation. Since it vanishes on every generator, it vanishes inductively on every polynomial in the fields. Hence $s^2\mathcal{O}=0$ for every such operator. This genuine result is stronger than the automatic vanishing obtained by merely setting ${ϵ}^2=0$; two independent transformation parameters also give a vanishing commutator.