Theorem: Selective visibility or erasure

For a fixed instrument and fixed $\sigma$, its complete output on ${\tau }_{\sigma }(z)$ is constant on a nonempty open interval containing zero if and only if

In that case it is constant on the entire declared real-amplitude family.

For the classical law alone, the corresponding condition is

Proof. Differentiate (eq:selective):

At zero, constancy forces $B_y=0$. At any nonzero point in the interval, $s(z)t(z)\ne 0$, so $C_y=0$. The converse follows from the exact formula. Distinct classical blocks cannot cancel. Taking traces gives the outcome-law statement.