= Solution
With the stated curvature convention, the <Ricci tensor> is $R_{bd}=R^a{}_{bad}$ and the <scalar curvature> is $R=g^{bd}R_{bd}$. Contract the differential <Bianchi identity> once to give the divergence of the <Riemann tensor>, then again with the inverse metric. The curvature antisymmetries yield $\nabla^aR_{ab}-\nabla_bR+\nabla^aR_{ab}=0$, so $\nabla^aR_{ab}=\tfrac12\nabla_bR$.
The <Einstein field equations> are $G_{ab}+\Lambda g_{ab}=(8\pi G/c^4)T_{ab}$, with $G_{ab}=R_{ab}-\tfrac12Rg_{ab}$. The <contracted Bianchi identity> makes the left side divergence-free, enforcing the local <stress-energy conservation> law $\nabla^aT_{ab}=0$ and ensuring consistency of the matter equations with the gravitational constraints.
For a scalar, second covariant derivatives commute. Commuting a further derivative past the resulting covector introduces <Ricci curvature>, giving $\nabla^2\nabla_a\phi=\nabla_a\nabla^2\phi+R_{ab}\nabla^b\phi$.
Now $R_{ab}=\nabla_a\nabla_b\phi$ implies $R=\nabla^2\phi$. Taking its divergence and using that commutation formula gives $\tfrac12\nabla_bR=\nabla_bR+R_{ba}\nabla^a\phi$, hence $\nabla_bR=-2R_{ba}\nabla^a\phi$. Meanwhile $\nabla_b(\nabla_a\phi\nabla^a\phi)=2R_{ba}\nabla^a\phi$. Adding proves
$$
\boxed{R+\nabla_a\phi\nabla^a\phi=\text{constant on each connected component}.}
$$
In Riemannian signature this is the <steady gradient Ricci soliton scalar identity>, with soliton potential $f=-\phi$.
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