Completion of the proof for Family 3.6

Proof of Theorem 103.1. Suppose that $\beta_X(F)\le0$ for a prime divisor $F$ over $X$. By Proposition 104.6, its centre is contained in a double conic $2\ell$. Take the weighted extraction $\sigma:Y\to X$ constructed above. It is plt, its different is $D/2$, and its actual refinement has threshold at least one by Proposition 106.10. Abban–Zhuang adjunction in its strict form gives $$\beta_X(F)\ge\operatorname{ord}_F(G) \left(A_X(G)-S_X(G)\right) =\frac{21}{88}\operatorname{ord}_F(G)>0.$$ The last inequality holds because $c_X(F)\subset\ell$, so the centre on $Y$ is contained in the exceptional divisor. This is the required contradiction.

For clarity, the strict inequality retains more information than the minimum bound for local delta invariants. Choose a basis compatible with the $F$- and $G$-filtrations and restrict its moving part to $G$. The refined bound and inversion of adjunction give, after taking limits, $$S_X(F)\le A_X(F)-A_X(G)\operatorname{ord}_F(G) +S_X(G)\operatorname{ord}_F(G).$$ This is precisely the displayed positive remainder. Thus a refined threshold equal to one causes no gap in the argument. ◻