Cohomology of the Koszul-Tate complex for an irreducible symmetry vanishes in degree $-2$

There must be something really obvious that I am missing here but any help is appreciated.

Suppose I have a theory with some action $S$ on some fields $phi$ such that any function vanishing on-shell has the form $frac{partial S}{partialphi^i}F^i$ for some $F^i$ functions of the fields. Assume that an irreducible complete set of gauge symmetries of $S$ is given by the functions $R^i_a$ of the fields. Completeness means that if some functions $lambda^i$ of the fields satisfy $frac{partial S}{partialphi^i}lambda^i=0$, then $lambda^i=R^i_a F^a+frac{partial S}{partialphi^j}F^{ij}$, for some functions $F^a$ and $F^{ij}=-F^{ji}$ of the fields. Irreducibility means that if a gauge transformation is trivial, that is, if we have functions $lambda^a$ and $mu^{ij}=-mu^{ji}$ of the fields such that $R^i_alambda^a=frac{partial S}{partialphi^j}mu^{ji}$, then there are functions $nu^{ai}$ of the fields such that $lambda^a=frac{partial S}{partial phi^j}nu^{aj}$.

Accourding to Quantization of Gauge Systems of Henneaux and Teitelboim, the Koszul-Tate complex is defined by adding a bosonic antifield to a ghost $c^*_a$ and a fermionic antifield $phi^*_i$, in degrees $-2$ and $-1$ respectively, along with a differential $delta$, which is the degree $1$ vector field defined by
$$deltaphi^i=0,quaddeltaphi^*_i=frac{partial S}{partialphi^i},quaddelta c^*_a=phi^*_iR^i_a.$$
It is claimed that the cohomology in degree $-2$ should vanish. I don’t see how this is the case.

A general closed element is of the form $c^*_aF^a+frac{1}{2}phi^*_iphi^*_jF^{ij}$, for functions $F^a$ and $F^{ij}=-F^{ji}$ of the fields, satisfying
$$0=deltaleft(c^*_aF^a+frac{1}{2}phi^*_iphi^*_jF^{ij}right)=phi^*_jleft(R^j_aF^a+frac{partial S}{partialphi^i}F^{ij}right),$$
i.e., such that the gauge transformation $R^j_aF^a$ turns out to be trivial and, in fact, equal to $-frac{partial S}{partialphi^i}F^{ij}$. What we want to show is that every such element is exact. This means that it can be written in the form
$$deltaleft(phi^*_iF^{ia}c^*_a+frac{1}{6}phi^*_iphi^*_jphi^*_kF^{ijk}right)=frac{partial S}{partialphi^i}F^{ia}c^*_a-phi^*_iphi^*_jF^{ia}R^j_a+frac{partial S}{partialphi^i}phi^*_jphi^*_kF^{ijk},$$
for some functions $F^{ia}$ and $F^{ijk}$ (completely antisymmetric) of the fields. Equivalently, we want to find such functions for which
$$F^a=frac{partial S}{partial phi^i}F^{ai},quad F^{ij}=F^{ja}R^i_a-F^{ia}R^j_a+2frac{partial S}{partialphi^k}F^{ijk}.$$

The existence of a function $F^{ai}$ satisfying the equation for $F^a$ is a direct consequence of irreducibility of the gauge transformations. However, I do not see how one can tune such a function to ensure that the second equation is also satisfied. Any help is much appreciated.

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So, I thought I solved it. From the irreducibility of the gauge transformations there is an $M^{ai}$ such that $F^a=frac{partial S}{partial phi^i}M^{ai}$. Then we have
$$R^j_afrac{partial S}{partial phi^i}M^{ai}=-frac{partial S}{partialphi^i}F^{ij}.$$
Then completeness of the gauge transformations guarantees the existence of $N^{ij}$ and $T^{kij}=-T^{ikj}$ such that ($star$)
$$R^j_aM^{ai}+F^{ij}=R^i_aN^{aj}+frac{partial S}{partial phi^k}T^{kij}.$$
Antisymmetrization with respect to $i$ and $j$ then yields
$$F^{ij}=F^{ja}R^i_a-F^{ia}R^j_a+2frac{partial S}{partialphi^k}F^{ijk},$$
with $F^{ia}=(N^{ia}+M^{ia})/2$ and $F^{ijk}=(T^{ijk}+T^{jik})/2$. The problem would then be solved if
$$frac{partial S}{partialphi^i}M^{ai}=F^a=frac{partial S}{partialphi^i}F^{ai}.$$
This would be true if
$$frac{partial S}{partialphi^i}N^{ai}=frac{partial S}{partialphi^i}M^{ai}.$$
However, I don’t see why this should happen.