Proposition: Action symmetry does not establish observable closure

An identity involving only an action and its differential does not by itself establish that a separate observable is closed under that differential.

Proof. The equation for the action contains no condition determining the differential of every other element. Algebraically, consider the graded polynomial algebra generated by an even element $x$ and an odd element $\xi$, with differential

Take the action element to be $S=0$ and the observable to be $O=x$. Then $dS=0$ but $dO=\xi \ne 0$. Thus closure of the action is not a sufficient abstract premise for closure of an independently specified observable. An additional observable theorem is required.