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An invariant path-integral measure and BRST symmetry action imply this identity when every separated insertion $G_i$ is BRST-closed. The bracket is the graded commutator, so it becomes an anticommutator for an odd $\mathcal{O}$. The arbitrary operator need not itself be closed. Exact insertions therefore decouple from physical correlators.

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.

Let $G_1,\dots ,G_r$ be the separated gauge-invariant insertions. Their BRST symmetry variations vanish: the gauge field strength and its gauge covariant derivatives transform by adjoint commutators, and invariant color contractions remove these commutators. Thus $s{G}_i=0$.

Make the infinitesimal change of integration variables $\Phi \mapsto \Phi +ϵs\Phi$ in the normalized expectation of $\mathcal{O}(x){\prod }_i{G}_i$. The action is invariant up to its boundary term and, by assumption, the measure has no Jacobian anomaly. ThereforeSince $s\mathcal{O}$ is generated by the BRST charge, this is the graded BRST Ward identityHere $[Q,\mathcal{O}\}=Q\mathcal{O}-(-1{)}^{|\mathcal{O}|}\mathcal{O}Q$ is the graded commutator. For even $\mathcal{O}$ it is the ordinary commutator printed in the question; for odd $\mathcal{O}$ it is an anticommutator. Separation of the other insertions from $x$ avoids the additional coincident-point contact terms. An operator need not itself be BRST-closed for this identity to hold.

A BRST-exact insertion has zero correlation with physical, gauge-invariant insertions. This is the decoupling of BRST-exact insertions in physical correlation functions. In the associated BRST cohomology, physical information is represented by closed states or operators modulo exact ones, schematically $\ker Q/\mathrm{im}Q$, using the nilpotence of the BRST charge in the anomaly-free theory. Gauge-fixing fields thereby do not supply additional physical observables.

For example, the gauge parameter dependence is exact:Differentiating a normalized physical correlation function with respect to $\xi$ inserts this expression; the BRST Ward identity makes that derivative zero under the same invariant-measure and boundary assumptions. This illustrates gauge-fixing parameter independence from BRST symmetry.