Every boundary atomic edge occurs in the square boundary chain

Let D be a SquareDissection: a natural number n, a finite vertex type with decidable equality and injective real-plane coordinates, and n nondegenerate vertex triples whose closed convex hulls cover exactly $Q=[0,1]^2$, have pairwise disjoint topological interiors, and each have area $1/n$ (the rational quotient embedded in the reals). Area is half the absolute determinant. Triangle sides are subdivided at all vertices lying strictly between their endpoints, ordered by affine parameter. Consecutive vertices form unordered atomic edges; multiplicity counts occurrences across the triangle boundary lists. T-junctions and unused vertices are permitted. No oddness assumption on n is made here.

Suppose vertices $c_{00},c_{10},c_{11},c_{01}$ have coordinates $(0,0),(1,0),(1,1),(0,1)$ respectively. Let L be the concatenation of the consecutive-edge lists along the four chains from $c_{00}$ to $c_{10}$ to $c_{11}$ to $c_{01}$ to $c_{00}$, each chain containing all D-vertices strictly between its endpoints in affine order.

If an unordered edge e occurs in some triangle atomic-edge list and its entire geometric segment lies in the frontier of Q, then e belongs to the finite set of entries of L. This is the inclusion from boundary atomic edges to the boundary chain.