Let $F \to E \to M$ be a smooth fiber bundle with connection $\omega$ and curvature $R$. We can form a (graded) vector bundle by taking the complex of differential forms at each fiber. Call this bundle $A(E) \to M$ where a fiber above $p\in M$ is $\Omega^*(E_p)$ i.e. differential forms on the fiber $E_p$. The connection on the fiber bundle induces a vector bundle connection on $A(E) \to M$, call it $\nabla$. We know that on a local trivialization $U$ we have $\nabla|_U \in \Omega^1(U, End( U \times \Omega^*(F))$. Also, we know that the curvature $R$ of the original fiber bundle is given by a $2$-form $R \in \Omega^2(U, \chi(F))$, i.e. a $2$-form on the base with values on vector fields of $F$.
I want to show that for any vector fields $X,Z,Y$ on $U$ we have that $\nabla|_U(X)\circ \iota_{R(Y,Z)}+ \iota_{R(X,Y)} \circ \nabla|_U(Z)=0$, where $\iota_{R(X,Y)}:\Omega^*(F)\to \Omega^{*-1}(F)$ is contraction with the vector field $R(X,Y)$.
Does this follow from the Bianchi identity?