Artin right complements satisfy the cube condition

For a letter $s$ and a word $w$ all of whose letters are distant from $s$ (that is, $\mathrm{\mid }i-j\mathrm{\mid }\ge 2$ for $s={\sigma }_i$ and each letter ${\sigma }_j$ of $w$), we have $\mathrm{\Theta }(s,w)=w$. Indeed, for $w=\epsilon$ this is [F1]; for $w=t{w}^{\mathrm{\prime }}$ with $t$ distant from $s$ we have $\theta (s,t)=t$ and $\theta (t,s)=s$ by [F1], whence $\mathrm{\Theta }(s,t{w}^{\mathrm{\prime }})=t\mathrm{\Theta }(s,w^{\mathrm{\prime }})=t{w}^{\mathrm{\prime }}$ by induction on $\mathrm{\mid }w\mathrm{\mid }$, which is legitimate because $\mathrm{\mid }{w}^{\mathrm{\prime }}\mathrm{\mid }<\mathrm{\mid }w\mathrm{\mid }$ for the length function of [L3].

For every word $x$ we have $\mathrm{\Theta }(x,x)=\epsilon$. Indeed, for $x=\epsilon$ this is [F1]; for $x=s{x}^{\mathrm{\prime }}$ with $s$ a letter we get from [F1] that $\mathrm{\Theta }(s,s{x}^{\mathrm{\prime }})=\theta (s,s)\mathrm{\Theta }(\theta (s,s),x^{\mathrm{\prime }})=\epsilon \cdot \mathrm{\Theta }(\epsilon ,x^{\mathrm{\prime }})=x^{\mathrm{\prime }}$, hence $\mathrm{\Theta }(s{x}^{\mathrm{\prime }},s{x}^{\mathrm{\prime }})=\mathrm{\Theta }(x^{\mathrm{\prime }},\mathrm{\Theta }(s,s{x}^{\mathrm{\prime }}))=\mathrm{\Theta }(x^{\mathrm{\prime }},x^{\mathrm{\prime }})=\epsilon$ by induction on $\mathrm{\mid }x\mathrm{\mid }$ with the length function of [L3].

For two letters ${\sigma }_i,{\sigma }_j$ we have $\mathrm{\Theta }({\sigma }_i,{\sigma }_j)=\theta ({\sigma }_i,{\sigma }_j)$ by [F1]; in particular $\mathrm{\Theta }$ of two distant letters is the second letter.

Repeated entries. (a) If $u=v$, then $\mathrm{\Theta }(u,v)=\mathrm{\Theta }(u,u)=\mathrm{\Theta }(v,u)$ and $\mathrm{\Theta }(u,w)=\mathrm{\Theta }(v,w)$, so ${\mathrm{\Theta }}_3(u,v,w)$ and ${\mathrm{\Theta }}_3(v,u,w)$ are the same word and are trivially equivalent. (b) If $u=w$, then $\mathrm{\Theta }(u,u)=\epsilon$ by step 1.2, so ${\mathrm{\Theta }}_3(u,v,w)=\mathrm{\Theta }(\mathrm{\Theta }(u,v),\epsilon )=\epsilon$ and ${\mathrm{\Theta }}_3(v,u,u)=\mathrm{\Theta }(\mathrm{\Theta }(v,u),\mathrm{\Theta }(v,u))=\epsilon$ by step 1.2; the two words are equal. (c) If $v=w$, then ${\mathrm{\Theta }}_3(u,v,v)=\mathrm{\Theta }(\mathrm{\Theta }(u,v),\mathrm{\Theta }(u,v))=\epsilon$ by step 1.2, and ${\mathrm{\Theta }}_3(v,u,v)=\mathrm{\Theta }(\mathrm{\Theta }(v,u),\mathrm{\Theta }(v,v))=\mathrm{\Theta }(\mathrm{\Theta }(v,u),\epsilon )=\epsilon$ by [F1] and step 1.2, so again the two words are equal. Hence the cube condition holds for every triple with a repeated entry.

Triples with no adjacent pair. Assume ${\sigma }_i,{\sigma }_j,{\sigma }_k$ are pairwise distant. Then $\mathrm{\Theta }({\sigma }_i,{\sigma }_j)={\sigma }_j$, $\mathrm{\Theta }({\sigma }_i,{\sigma }_k)={\sigma }_k$ by step 1.3, and ${\sigma }_j,{\sigma }_k$ are distant, so ${\mathrm{\Theta }}_3({\sigma }_i,{\sigma }_j,{\sigma }_k)=\mathrm{\Theta }({\sigma }_j,{\sigma }_k)={\sigma }_k$; likewise $\mathrm{\Theta }({\sigma }_j,{\sigma }_i)={\sigma }_i$, $\mathrm{\Theta }({\sigma }_j,{\sigma }_k)={\sigma }_k$, so ${\mathrm{\Theta }}_3({\sigma }_j,{\sigma }_i,{\sigma }_k)=\mathrm{\Theta }({\sigma }_i,{\sigma }_k)={\sigma }_k$. The two sides are equal.

Triples with exactly one adjacent pair. Let ${\sigma }_i,{\sigma }_{i+1},{\sigma }_k$ with ${\sigma }_k$ distant from both ${\sigma }_i$ and ${\sigma }_{i+1}$, that is $k\mathrm{\notin }\{i-1,i,i+1,i+2\}$. Then, using [F1] and step 1.3, ${\mathrm{\Theta }}_3({\sigma }_i,{\sigma }_{i+1},{\sigma }_k)=\mathrm{\Theta }({\sigma }_{i+1}{\sigma }_i,\mathrm{\Theta }({\sigma }_i,{\sigma }_k))=\mathrm{\Theta }({\sigma }_{i+1}{\sigma }_i,{\sigma }_k)=\mathrm{\Theta }({\sigma }_i,\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_k))=\mathrm{\Theta }({\sigma }_i,{\sigma }_k)={\sigma }_k,$ and ${\mathrm{\Theta }}_3({\sigma }_{i+1},{\sigma }_i,{\sigma }_k)=\mathrm{\Theta }({\sigma }_i{\sigma }_{i+1},\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_k))=\mathrm{\Theta }({\sigma }_i{\sigma }_{i+1},{\sigma }_k)=\mathrm{\Theta }({\sigma }_{i+1},\mathrm{\Theta }({\sigma }_i,{\sigma }_k))=\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_k)={\sigma }_k\mathrm{.}$ The two sides are equal; the identity $\mathrm{\Theta }({\sigma }_{i+1}{\sigma }_i,{\sigma }_k)=\mathrm{\Theta }({\sigma }_i,\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_k))$ is the defining recursion, and $\mathrm{\Theta }({\sigma }_i,{\sigma }_k)={\sigma }_k$, $\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_k)={\sigma }_k$ hold because $k$ is distant from $i$ and from $i+1$. To cover the other placements, write $a:={\sigma }_i$, $b:={\sigma }_{i+1}$ and $c:={\sigma }_k$. If the adjacent pair occupies the first and third positions, then ${\mathrm{\Theta }}_3(a,c,b)=\mathrm{\Theta }(c,ba)=ba$ by step 1.1, while ${\mathrm{\Theta }}_3(c,a,b)=\mathrm{\Theta }(a,b)=ba$ by step 1.3 and [F1]. Interchanging the names $a,b$ gives ${\mathrm{\Theta }}_3(b,c,a)=\mathrm{\Theta }(c,ab)=ab=\mathrm{\Theta }(b,a)={\mathrm{\Theta }}_3(c,b,a)$. These two equalities and the equality with $w=c$ already computed cover all six orders of the three distinct letters; swapping the first two arguments merely reverses one of these equalities.

Three consecutive indices, first case. Let $1\le i\le n-3$. Using [F1] and the values $\mathrm{\Theta }({\sigma }_i,{\sigma }_{i+1})={\sigma }_{i+1}{\sigma }_i$, $\mathrm{\Theta }({\sigma }_i,{\sigma }_{i+2})={\sigma }_{i+2}$ (indices differing by $2$), ${\mathrm{\Theta }}_3({\sigma }_i,{\sigma }_{i+1},{\sigma }_{i+2})=\mathrm{\Theta }({\sigma }_{i+1}{\sigma }_i,{\sigma }_{i+2})=\mathrm{\Theta }({\sigma }_i,\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_{i+2}))=\mathrm{\Theta }({\sigma }_i,{\sigma }_{i+2}{\sigma }_{i+1})=\theta ({\sigma }_i,{\sigma }_{i+2})\mathrm{\Theta }(\theta ({\sigma }_{i+2},{\sigma }_i),{\sigma }_{i+1})={\sigma }_{i+2}\mathrm{\Theta }({\sigma }_i,{\sigma }_{i+1})={\sigma }_{i+2}{\sigma }_{i+1}{\sigma }_i\mathrm{.}$ For the second side, $\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_i)={\sigma }_i{\sigma }_{i+1}$ and $\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_{i+2})={\sigma }_{i+2}{\sigma }_{i+1}$, so ${\mathrm{\Theta }}_3({\sigma }_{i+1},{\sigma }_i,{\sigma }_{i+2})=\mathrm{\Theta }({\sigma }_i{\sigma }_{i+1},{\sigma }_{i+2}{\sigma }_{i+1})=\mathrm{\Theta }({\sigma }_{i+1},\mathrm{\Theta }({\sigma }_i,{\sigma }_{i+2}{\sigma }_{i+1}))=\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_{i+2}{\sigma }_{i+1}{\sigma }_i)={\sigma }_{i+2}{\sigma }_{i+1}\mathrm{\Theta }({\sigma }_{i+1}{\sigma }_{i+2},{\sigma }_{i+1}{\sigma }_i),$ and $\mathrm{\Theta }({\sigma }_{i+1}{\sigma }_{i+2},{\sigma }_{i+1}{\sigma }_i)=\mathrm{\Theta }({\sigma }_{i+2},\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_{i+1}{\sigma }_i))=\mathrm{\Theta }({\sigma }_{i+2},{\sigma }_i)={\sigma }_i$, since $\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_{i+1}{\sigma }_i)={\sigma }_i$; hence the second side is ${\sigma }_{i+2}{\sigma }_{i+1}{\sigma }_i$ as well, and the two sides are equal.

Three consecutive indices, second case. Here $\mathrm{\Theta }({\sigma }_{i+2},{\sigma }_{i+1})={\sigma }_{i+1}{\sigma }_{i+2}$, $\mathrm{\Theta }({\sigma }_{i+2},{\sigma }_i)={\sigma }_i$, so ${\mathrm{\Theta }}_3({\sigma }_{i+2},{\sigma }_{i+1},{\sigma }_i)=\mathrm{\Theta }({\sigma }_{i+1}{\sigma }_{i+2},{\sigma }_i)=\mathrm{\Theta }({\sigma }_{i+2},\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_i))=\mathrm{\Theta }({\sigma }_{i+2},{\sigma }_i{\sigma }_{i+1})=\theta ({\sigma }_{i+2},{\sigma }_i)\mathrm{\Theta }(\theta ({\sigma }_i,{\sigma }_{i+2}),{\sigma }_{i+1})={\sigma }_i\mathrm{\Theta }({\sigma }_{i+2},{\sigma }_{i+1})={\sigma }_i{\sigma }_{i+1}{\sigma }_{i+2}\mathrm{.}$ For the other side, $\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_{i+2})={\sigma }_{i+2}{\sigma }_{i+1}$ and $\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_i)={\sigma }_i{\sigma }_{i+1}$, so ${\mathrm{\Theta }}_3({\sigma }_{i+1},{\sigma }_{i+2},{\sigma }_i)=\mathrm{\Theta }({\sigma }_{i+2}{\sigma }_{i+1},{\sigma }_i{\sigma }_{i+1})=\mathrm{\Theta }({\sigma }_{i+1},\mathrm{\Theta }({\sigma }_{i+2},{\sigma }_i{\sigma }_{i+1}))=\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_i{\sigma }_{i+1}{\sigma }_{i+2}),$ where $\mathrm{\Theta }({\sigma }_{i+2},{\sigma }_i{\sigma }_{i+1})=\theta ({\sigma }_{i+2},{\sigma }_i)\mathrm{\Theta }(\theta ({\sigma }_i,{\sigma }_{i+2}),{\sigma }_{i+1})={\sigma }_i\mathrm{\Theta }({\sigma }_{i+2},{\sigma }_{i+1})={\sigma }_i{\sigma }_{i+1}{\sigma }_{i+2}$; continuing, $\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_i{\sigma }_{i+1}{\sigma }_{i+2})=\theta ({\sigma }_{i+1},{\sigma }_i)\mathrm{\Theta }(\theta ({\sigma }_i,{\sigma }_{i+1}),{\sigma }_{i+1}{\sigma }_{i+2})={\sigma }_i{\sigma }_{i+1}\mathrm{\Theta }({\sigma }_{i+1}{\sigma }_i,{\sigma }_{i+1}{\sigma }_{i+2})$, and $\mathrm{\Theta }({\sigma }_{i+1}{\sigma }_i,{\sigma }_{i+1}{\sigma }_{i+2})=\mathrm{\Theta }({\sigma }_i,\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_{i+1}{\sigma }_{i+2}))=\mathrm{\Theta }({\sigma }_i,{\sigma }_{i+2})={\sigma }_{i+2}$, because $\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_{i+1}{\sigma }_{i+2})={\sigma }_{i+2}$. Hence ${\mathrm{\Theta }}_3({\sigma }_{i+1},{\sigma }_{i+2},{\sigma }_i)={\sigma }_i{\sigma }_{i+1}{\sigma }_{i+2}$, equal to the first side.

Three consecutive indices, third case. ${\mathrm{\Theta }}_3({\sigma }_{i+2},{\sigma }_i,{\sigma }_{i+1})=\mathrm{\Theta }(\mathrm{\Theta }({\sigma }_{i+2},{\sigma }_i),\mathrm{\Theta }({\sigma }_{i+2},{\sigma }_{i+1}))=\mathrm{\Theta }({\sigma }_i,{\sigma }_{i+1}{\sigma }_{i+2})=\theta ({\sigma }_i,{\sigma }_{i+1})\mathrm{\Theta }(\theta ({\sigma }_{i+1},{\sigma }_i),{\sigma }_{i+2})={\sigma }_{i+1}{\sigma }_i\mathrm{\Theta }({\sigma }_i{\sigma }_{i+1},{\sigma }_{i+2}),$ and $\mathrm{\Theta }({\sigma }_i{\sigma }_{i+1},{\sigma }_{i+2})=\mathrm{\Theta }({\sigma }_{i+1},\mathrm{\Theta }({\sigma }_i,{\sigma }_{i+2}))=\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_{i+2})={\sigma }_{i+2}{\sigma }_{i+1}$, so ${\mathrm{\Theta }}_3({\sigma }_{i+2},{\sigma }_i,{\sigma }_{i+1})={\sigma }_{i+1}{\sigma }_i{\sigma }_{i+2}{\sigma }_{i+1}$. Likewise ${\mathrm{\Theta }}_3({\sigma }_i,{\sigma }_{i+2},{\sigma }_{i+1})=\mathrm{\Theta }(\mathrm{\Theta }({\sigma }_i,{\sigma }_{i+2}),\mathrm{\Theta }({\sigma }_i,{\sigma }_{i+1}))=\mathrm{\Theta }({\sigma }_{i+2},{\sigma }_{i+1}{\sigma }_i)=\theta ({\sigma }_{i+2},{\sigma }_{i+1})\mathrm{\Theta }(\theta ({\sigma }_{i+1},{\sigma }_{i+2}),{\sigma }_i)=({\sigma }_{i+1}{\sigma }_{i+2})\mathrm{\Theta }({\sigma }_{i+2}{\sigma }_{i+1},{\sigma }_i)={\sigma }_{i+1}{\sigma }_{i+2}\mathrm{\Theta }({\sigma }_{i+1},\mathrm{\Theta }({\sigma }_{i+2},{\sigma }_i))={\sigma }_{i+1}{\sigma }_{i+2}\mathrm{\Theta }({\sigma }_{i+1},{\sigma }_i)={\sigma }_{i+1}{\sigma }_{i+2}{\sigma }_i{\sigma }_{i+1}$. The two words differ only in the order of the distant letters ${\sigma }_i$ and ${\sigma }_{i+2}$, so they are ${\equiv }^{+}$-equivalent by [F2].

Enumerating the patterns. Let $u,v,w$ be letters with $u,v,w$ pairwise distinct, and consider the graph on the three indices with an edge for each adjacent pair. It has at most two edges, since $\{1,\dots ,n-1\}$ with the adjacency relation is a path and a path has no triangle; if it has no edge, step 2.2 applies; if it has exactly one edge, step 2.3 covers all six orders; if it has two edges, the three indices are $i,i+1,i+2$ in some order and steps 2.4--2.6 cover three orders, and swapping the first two arguments covers the other three. Together with the repeated-entry case of step 2.1 this covers every triple of letters.

Every triple $(u,v,w)$ of letters therefore satisfies ${\mathrm{\Theta }}_3(u,v,w){\equiv }^{+}{\mathrm{\Theta }}_3(v,u,w)$, which is the $\theta$-cube condition for generators; the displayed values of the statement are steps 2.4--2.6. ∎