Free Cayley trees from reduced-word normal form

A reduced word for $g^{-1}h$ gives a finite path from $g$ to $h$ by successively multiplying on the right by its letters, so the graph is connected. On an edge from $g_i$ to $g_{i+1}$ its right label is $s_i=g_i^{-1}{g}_{i+1}$, a formal basis letter or inverse. Distinct formal letters represent distinct nonidentity elements by F1, so there are no loops or parallel edges after reversal identification.

A simple cycle would have no immediate reversal, so its successive labels form a nonempty reduced word $s_0\cdots s_{n-1}$. But their ordered product telescopes to $g_0^{-1}{g}_n=e$, contradicting F1, since the identity is the empty reduced word. This also excludes two-edge reversal-free cycles; loops were already excluded. Thus the graph has no cycles, and F2 makes its realization a geodesic tree with tripod triangles.

A finite edge path with no immediate reversal cannot repeat a vertex: the labels along any closed portion would again form a nonempty reduced identity word. Hence it is the unique simple path between its endpoints. F2 makes its length parametrization geodesic. For a real subinterval of an edge path, bracket its endpoints by adjacent vertices, using the endpoints themselves if already vertices. A finite bracketing path is simple by the same argument, so its entire geometric realization and every restriction are isometric. For a path ending inside an edge, the same conclusion follows by subdivision at that endpoint. This treats finite, one-sided infinite and two-sided infinite paths because each compact restriction meets finitely many unit edges.

The empty path and one-point real intervals have distance and length zero. If $S$ is empty there is only the empty reduced word, no edges and the one-point realization, so all assertions still hold. No family of paths or representatives has been selected; the proof is choice-free.