and cannot be applied to the first word. This leads to a contradiction, which proves this property.

This condition also implies the property (2) for $\varphi$. For this we need to compute the lcm’s between two elements $x,y$ of $S$. We need to prove the following in $M(S_0)$, for any $S_0 \subset S$ satisfying the above condition that contains $x$ and $y$.

-   $\lcm(s_i,s_j) = s_i s_j = s_j s_i$ if $|j-i| \geq 2$

-   $\lcm(s_i,s_{i+1}) = s_i s_{i+1} s_i = s_{i+1} s_i s_{i+1}$

-   $\lcm(t_i,t_j) = t_1 t_0 = t_i t_{i-1} = t_jt_{j-1}$

-   $\lcm(t_i ,s_3) = t_i s_3 t_i = s_3 t_i s_3 $

-   $\lcm(t_i, s_j) = t_is_j = s_j t_i $ if $j \geq 4$.

The identities with length two are clear, as the lcm exist and cannot have length 1. For the ones of length 3, namely $\{ x, y \} = \{ s_i,s_{i+1} \}$ and $\{ x, y \}  =  \{ t_i ,s_3 \}$, we use that $\{ x , y \} \subset S_0 \subset S$ satisfies our condition. Since the lcm of $x,y$ in $M(S_0)$ should divide $xyx=yxy$, it should then come from $M(\{ x, y \})$, meaning that is should be a 