In a reduced C prime(1/6) null diagram, some face contributes more than half of its boundary to the outer boundary

If $D$ had no faces, [F1] would make it a tree. Every nontrivial closed walk in a tree has immediate backtracking, contradicting the freely reduced nontrivial boundary word. Thus $D$ has a face, and by [L1] some boundary face $f$ of $D$ is a shell whose inner boundary is a concatenation $q_1\cdots q_i$ of maximal internal arcs with $0\le i\le 3$. (For $i=0$, this is the empty concatenation.) Let $p$ be the complementary outer arc of $\mathrm{\partial }f$ lying on $\mathrm{\partial }D$.

If $i=0$, the sum of the inner-arc lengths is $0<\mathrm{\mid }\mathrm{\partial }f\mathrm{\mid }\mathrm{/}2$. If $1\le i\le 3$, each internal arc $q_j$ is a piece by [L2], including a possible self-incident arc. The strict $C^{\mathrm{\prime }}(1\mathrm{/}6)$ bound in [L2] gives $\mathrm{\mid }{q}_j\mathrm{\mid }<\mathrm{\mid }\mathrm{\partial }f\mathrm{\mid }\mathrm{/}6$, and hence $\mathrm{\mid }{q}_1\mathrm{\mid }+\cdots +\mathrm{\mid }{q}_i\mathrm{\mid }<i\frac{\mathrm{\mid }\mathrm{\partial }f\mathrm{\mid }}{6}\le \frac{\mathrm{\mid }\mathrm{\partial }f\mathrm{\mid }}{2}\mathrm{.}$ Thus in every case the total inner-arc length is less than half of $\mathrm{\mid }\mathrm{\partial }f\mathrm{\mid }$.

Since $\mathrm{\partial }f$ is the disjoint union of the outer arc $p$ and the inner arcs $q_1,\dots ,q_i$, step 2.1 gives $\mathrm{\mid }p\mathrm{\mid }=\mathrm{\mid }\mathrm{\partial }f\mathrm{\mid }-(\mathrm{\mid }{q}_1\mathrm{\mid }+\cdots +\mathrm{\mid }{q}_i\mathrm{\mid })>\frac{\mathrm{\mid }\mathrm{\partial }f\mathrm{\mid }}{2}\mathrm{.}$ Thus $f$ contributes more than half of its boundary to the outer boundary of $D$.