Proposition: A linear bridge does not admit an excluded coefficient

Let $\mathcal{P}$ be an off-shell polynomial domain, and let $\mathcal{B}$ be a functional with $\mathcal{B}\notin \mathcal{P}$. Suppose an admissible quantum algebra requires the classical coefficient of every admitted element to belong to $\mathcal{P}$, as an actual functional on the full off-shell configuration space.

Then no admitted element can have classical coefficient equal to $\mathcal{B}$ on that full space. A coefficientwise formal construction requiring $\mathcal{B}$ as an input cannot be repaired merely by proving an on-shell equality for a linear functional.

Proof. Such an admitted element would have a classical coefficient in $\mathcal{P}$ equal to $\mathcal{B}$, contradicting $\mathcal{B}\notin \mathcal{P}$.

An on-shell linear identity supplies neither membership of the bilinear functional in $\mathcal{P}$ nor equality of off-shell classical coefficients. It therefore does not remove the contradiction.