Proposition: Stress identities do not imply Einstein dynamics

Symmetry, conservation, and zero trace of a tensor on a prescribed metric do not imply Einstein's equation.

Proof. In a regular developed chart choose a nonzero constant null covector $k$ and put $t_{\mu \nu }=k_{\mu }{k}_{\nu }$. It is symmetric, constant and conserved, nonzero, and traceless. The local Einstein tensor is zero. Thus $G_{\mu \nu }=8\pi G_{\mathrm{N}}{t}_{\mu \nu }$ fails when $G_{\mathrm{N}}\ne 0$. No statement about a distributional stress tensor on $\mathcal{W}$ is needed.