Internal arcs of a reduced small cancellation diagram are pieces

Start each incident face-side reading at the initial endpoint of $a$, choosing its orientation to traverse $a$ in the given direction. The two readings have opposite planar orientations, even if they belong to the same face. Their words $r,s$ belong to the symmetrised set and both begin with the nonempty label $p$ of $a$. No free cancellation occurs inside $p$ since it is a segment of a cyclically reduced relator.

No nonempty cyclically reduced word equals a cyclic rotation of its inverse. Indeed, writing its letters $b_i$ with indices modulo its length $m$, such equality would give $b_i=b_{k-i}^{-1}$ for some integer $k$. If $m$ is odd, or if $m$ and $k$ are both even, $2i=k$ has a solution modulo $m$, forcing a letter to equal its formal inverse, impossible in $X^{\pm 1}$. If $m$ is even and $k$ is odd, take $i=(k-1)\mathrm{/}2$ modulo $m$; then $b_i=b_{i+1}^{-1}$, contrary to cyclic reduction. This includes proper-power words and uses no hypothesis on the presented group's torsion.

For distinct faces, equality $r=s$ would give a cancellable pair, excluded by [F2]. For the same face, the two opposite orientations make $s$ a cyclic rotation of $r^{-1}$, so step 1.2 excludes equality. Thus in both cases $r\mathrm{\ne }s$ and $p$ is a piece by [F1]. Under $C^{\mathrm{\prime }}(1\mathrm{/}6)$, apply its bound to $r$ and separately to $s$ to obtain $\mathrm{\mid }p\mathrm{\mid }<\mathrm{\mid }r\mathrm{\mid }\mathrm{/}6$ and $\mathrm{\mid }p\mathrm{\mid }<\mathrm{\mid }s\mathrm{\mid }\mathrm{/}6$. No injectivity of a face attaching walk has been assumed.