= Solution
The <Greendlinger lemma> says that a nonempty freely reduced <null-homotopic> combinatorial loop in a complex satisfying the <metric small cancellation condition> $C'(1/6)$ contains a consecutive segment of a cell boundary longer than half that boundary. In a reduced <disc diagram> without boundary spurs this is supplied by a boundary <disc diagram shell>, or by the single-cell case. The long segment can be replaced by the strictly shorter complementary segment.
The useful precise diagram version also allows degenerate diagrams. A <reduced disc diagram> is either a single vertex, a single closed cell, a <disc diagram ladder>, or has at least three <disc diagram spurs> or shells whose inner paths consist of at most three pieces. This is the <Greendlinger ladder theorem>. For such a shell, writing the perimeter as $QP$, where $Q$ is the exterior arc and $P$ is the union of its interior pieces, the strict $C'(1/6)$ inequalities give
$$
\boxed{|P|<\tfrac12|\partial R|,\qquad |Q|>\tfrac12|\partial R|.}
$$
A ladder is a chain of cells and possibly connecting edges, with the two boundary paths running along its opposite sides. In a nontrivial ladder with at least two blocks its two ends are shells or spurs. The single-cell case also has the long exterior arc conclusion when viewed against an appropriate boundary decomposition.
A tree diagram with no two-cells should not be described as containing a shell: it has a spur unless it is a point. Retaining the spur and ladder alternatives is essential for the quasiconvexity argument below.
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