The finite evaluation-span dual in full
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Fix a positive \(v\) and integer \(d\ge0\). Let
The constant monomial \(S=\varnothing\) is included. Degree-\(d\) coefficient covectors form the finite dual of this evaluation space. A covector is safe exactly when it annihilates \(N\).
If \(v_d(v)\in N\), every safe covector vanishes at \(v\). If \(v_d(v)\notin N\), extend a basis of \(N\) by \(v_d(v)\) and then to a basis of the full space. Assign zero to the \(N\) basis and one to \(v_d(v)\). The resulting covector is a safe detector. Hence
Choose a basis of \(N\) from negative evaluation rows. A span representation of the positive row gives a subset \(W\subseteq U\) with
It is a replayable finite obstruction to every safe degree-\(d\) detector.
At \(d=0\), a negative exists exactly when the constant-one detector is unsafe. At \(d\ge n\), the full evaluation rows are independent and no positive row lies in the negative span. This matches \(\beta_v\le n\) from point indicators. For \(f=0\), there is no positive \(v\) and the separate convention is \(\beta(0)=0\).
Fresh enumeration checked all 278 functions on zero through three inputs, all positives, and every \(d=0,\ldots,n\): 4,201 primal/dual checks agree. Those bounded checks support the independently supplied universal proof.