Enhanced A-polynomial Newton edges give strict boundary slopes

Let $e$ be any edge of the enhanced Newton polygon. By [F2], choose its minimum-normal valued model using the irreducible-character incidence charts in that supplier, not merely a factor of the ordinary polynomial. The finite word-matrix determinant certificate is inverted before the generic-fiber closed point is selected; hence it stays nonzero in the resulting finite field. Thus the constructed representation is absolutely irreducible, its peripheral eigenvalues are the actual coordinate functions, and their valuation pair is a positive multiple of the primitive normal to $e$. This also treats a factor $L-1$ contributed by irreducible characters.

Apply [F3] in the exterior verified by [F1]. It yields a connected orientable essential surface with nonempty boundary and the primitive kernel slope of that nonzero valuation pair. That is the edge slope by the same orthogonality equation as in [F4]. Moreover the last clause of [F3] applies to the ACTUAL absolutely irreducible representation from step 1.1. If the selected component were a fiber, its normal subgroup would fix an edge; integrality of all stable-letter conjugates would force the representation to be reducible, the contradiction proved there. Thus this surface is not a fiber of any fibration having it as fiber.

The surface is precisely a realizing orientable incompressible boundary-incompressible nonfiber surface in [F4]'s definition of strict boundary slope. Hence every specified enhanced edge slope is strict, including zero or infinity when those occur. No claim that the ordinary abelian $L-1$ edge is strict has been used: enhanced strictness depends on the actual irreducible-character model, while the ordinary boundary-slope theorem retains its weaker valid conclusion.