Every class in $W(X)/{\sim}$ 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 $|u|=n$; 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 $\Lambda_r=\Lambda_s$.

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

Applying the equal permutations of step 1.3 to the empty word gives $r=\Lambda_r(\varepsilon)=\Lambda_s(\varepsilon)=s$, because the construction in [L1] recovers every reduced word from $\varepsilon$.

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.