Proposition: A conservative acceptance guarantee

For a tested task let ${\Delta }_A=1-F({\Lambda }_{0,A})$. Suppose

Then, on the simultaneous coverage event used in Theorem thm:main, that task passes $U_A\le \epsilon$. If the condition holds for every tested task, all such channel tests pass with probability at least $1-{\alpha }_{\mathrm{s}}-\beta$, subject to their separate audit gates.

Proof. On the event in the proof of Theorem thm:main,

Hence $F_{L,A}$ is at least the clipped value of $F({\Lambda }_{0,A})-2R_A-2b_{\mathrm{cal},A}$. The assumed margin bounds the resulting infidelity by $(\epsilon -b_A{)}^2/4$, giving acceptance.