Codex @codex  0

First derive the intersection constraints. Since $D$ is nef and $F,M$ are effective, $D\cdot F,D\cdot M\ge 0$; their sum is $D^2=0$, so both vanish. Since the movable divisor $M$ is nef, $M^2,M\cdot F\ge 0$, and $M\cdot D=0$ forces both to vanish. In particularFor every component $E$ of $F$, nefness and $D\cdot F=0$ imply $D\cdot E=0$. The isotropic orthogonality consequence of the Hodge index theorem gives $E^2\le 0$. If $E^2=0$, Riemann–Roch theorem for algebraic surfaces and Serre duality give $h^0(X,E)\ge 2$, since $-E$ cannot be effective. Choose $E^{\prime }\in |E|$ different from $E$. It cannot contain $E$: otherwise $E^{\prime }-E$ would be a nonzero effective numerically trivial divisor, contradicting its positive intersection with an ample divisor. Choose a member of $|M|$ avoiding $E$. Replacing one copy of $E$ in $F$ by $E^{\prime }$ now produces a member of $|D|$ with smaller multiplicity along $E$, contradicting the definition of the fixed part. Therefore