If no product of two members of a generating set is the identity and the Cayley graph is a tree, the set is a free basis

Given: The hypotheses of the Statement.

The Cayley graph of a group $G$ with respect to a subset $S$ has vertex set $G$ and edge set $\{\{g,gs\}:g\in G,s\in (S\cup S^{-1})\setminus \{e\}\}$ (The Cayley graph of a group with respect to a subset).

A cycle is a closed walk of length at least three with distinct vertices apart from its endpoints; a forest is a simple graph with no cycle and a tree is a connected forest (Cycles, trees and forests in a simple graph on an arbitrary vertex set).

A simple graph is a tree if and only if every two of its vertices are joined by exactly one path (A nonempty simple graph is a tree if and only if each pair of vertices is joined by exactly one path).

The Cayley graph of a free group with respect to a free basis is a tree (The Cayley graph of a free group with respect to a free basis is a tree).

A free group on a set $X$ is a group $F(X)$ together with a map $i:X\to F(X)$ such that, for every group $G$ and every function $u:X\to G$, there is a unique group homomorphism $\widehat{u}:F(X)\to G$ satisfying (Free group on a set of generators).

The subset $B$ is a free basis of $F$ if $(F,i)$ is a free group on the set $B$ in the sense of. (A free basis of a group).

The reduced words on $X\bigsqcup X^{-1}$ form a group when the product of reduced words is their concatenation followed by free reduction. (Reduced words form the free group on an alphabet).

Every class in $W(X)\mathrm{/}\sim$ contains exactly one reduced word. (Every class in $W(X)\mathrm{/}\sim$ contains exactly one reduced word).

An elementary cancellation deletes two adjacent letters $x{x}^{-1}$ or $x^{-1}x$. A word is reduced if no elementary cancellation applies. (Words in an alphabet with formal inverses, elementary cancellation, and reduced words).