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

The empty word is reduced and freely equivalent to itself, establishing the existence claim for words of length zero.

Assume every word of length $n$ is freely equivalent to a reduced word, and write a word of length $n+1$ as $ua$ with $\mathrm{\mid }u\mathrm{\mid }=n$; by the induction hypothesis, $u\sim r$ for some reduced $r$, so the congruence property of [L3], applied with the one-letter word $a$ on the right, gives $ua\sim ra$.

If reduced words $r$ and $s$ lie in the same class, then $r\sim s$, so [L1] gives ${\mathrm{\Lambda }}_r={\mathrm{\Lambda }}_s$.

If $r$ is empty or its last letter is not $a^{-1}$, then $ra$ is reduced; otherwise $r=r^{\mathrm{\prime }}{a}^{-1}$ and one elementary cancellation carries $ra=r^{\mathrm{\prime }}{a}^{-1}a$ to the reduced word $r^{\mathrm{\prime }}$. Thus every word is freely equivalent to a reduced word.

Applying the equal permutations of step 1.3 to the empty word gives $r={\mathrm{\Lambda }}_r(\epsilon )={\mathrm{\Lambda }}_s(\epsilon )=s$, because the construction in [L1] recovers every reduced word from $\epsilon$.

Consequently every class $[w]$ contains at least one reduced representative.

Step 3.1 gives existence and step 2.2 gives uniqueness, so each class contains exactly one reduced word.