Greendlinger shell existence from the curvature count

The diagram exists and is reduced by [F3]. Root its finite block tree at the boundary basepoint (or the block containing that point). A terminal bridge away from the root gives a spur excursion wholly inside the linear word, hence consecutive inverse letters, impossible since $w$ is freely reduced. If there are no faces, the diagram is a tree; any nontrivial finite rooted tree has such a terminal spur away from the root. Since $w$ is nonempty, the diagram is not a point. There is therefore a terminal disc block, with no attachments away from its parent attachment. When the root block is the only block, it is a disc and the chosen basepoint is the sole point to avoid inside the exterior arc.

For a multi-face terminal block, choose one of the two shells supplied by [F1] whose exterior arc does not contain the attachment in its interior. If this is the root disc, avoid the basepoint instead. At most one of the two distinct exterior arcs can contain that specified point internally. The chosen arc therefore appears contiguously in the original outer boundary walk. For a one-face terminal block, the entire face boundary is a contiguous excursion from the attachment back to itself; when it is the root disc, start the full face reading at the given basepoint, so it is the original linear word. This also handles conjugating bridges leading from the basepoint to a single disc.

For a shell with $i\in \{1,2,3\}$ internal arcs, their total length $t$ satisfies $t<i\mathrm{\mid }r\mathrm{\mid }\mathrm{/}6\le \mathrm{\mid }r\mathrm{\mid }\mathrm{/}2$ by [F2]. Its exterior arc has length $\mathrm{\mid }r\mathrm{\mid }-t>\mathrm{\mid }r\mathrm{\mid }\mathrm{/}2$. For a zero-shell it has length $\mathrm{\mid }r\mathrm{\mid }>\mathrm{\mid }r\mathrm{\mid }\mathrm{/}2$, since relators are nonempty. Step 2.1 places this segment in the original linear word. Choose the orientation and cyclic conjugate of the face relator that starts with this segment; symmetrisation ensures that this is again a defining relator. This proves the stated initial-segment conclusion.