The right weak interval below the longest element of $A_2$ is not distributive

The group and its elements. Since $S=\{s_1,s_2\}$, the subgroup $⟨s_1,s_2⟩$ is all of $W$, so by [F4] $W$ is dihedral of order $2\cdot 3=6$; its elements are the six distinct elements $(s_1{s}_2{)}^k$ and $(s_1{s}_2{)}^k{s}_1$ for $k=0,1,2$. With $(s_1{s}_2{)}^3=1$ from the relator of [F1] these are $1$, $s_1{s}_2$, $(s_1{s}_2{)}^2=s_2{s}_1$, $s_1$, $(s_1{s}_2)s_1=s_1{s}_2{s}_1=w_0$ and $(s_1{s}_2{)}^2{s}_1=s_2$; hence $W=\{1,s_1,s_2,s_1{s}_2,s_2{s}_1,w_0\}$ and, since the displayed words are the reduced expressions by [F4], the lengths are $0,1,1,2,2,3$ respectively, in particular $\mathrm{\ell }(w_0)=3$. The reduced words of $w_0$ are exactly $s_1{s}_2{s}_1$ and $s_2{s}_1{s}_2$: both are reduced of length $3$ and represent $w_0$ by the braid relation of [F1], and by F4 every reduced word of $w_0$ is braid-equivalent to $s_1{s}_2{s}_1$, while the only braid move applicable to a length-three word in the two letters replaces the whole alternating word by the other.

The interval and the covers. By the prefix property F2, the elements of $[1,w_0{]}_R$ are the products of the prefixes of the reduced words $s_1{s}_2{s}_1$ and $s_2{s}_1{s}_2$ of $w_0$ established in step 1.1, namely $1,s_1,s_1{s}_2,w_0$ and $1,s_2,s_2{s}_1,w_0$; these are all six elements of $W$ by 1.1, so $[1,w_0{]}_R=W$. By [F3] every cover in ${\le }_R$ is of the form $v=us$ with $s\in S$ and $\mathrm{\ell }(v)=\mathrm{\ell }(u)+1$; running over the six elements $u$ and the two generators and using the length table of 1.1, the products with length increase one are exactly $1\cdot s_1=s_1$, $1\cdot s_2=s_2$, $s_1\cdot s_2=s_1{s}_2$, $s_2\cdot s_1=s_2{s}_1$, $s_1{s}_2\cdot s_1=w_0$ and $s_2{s}_1\cdot s_2=w_0$, while $s_1{s}_2\cdot s_2=s_1$, $s_2{s}_1\cdot s_1=s_2$ and the products $w_{0}s$ (of length $2$) do not raise the length. Hence these six pairs are exactly the covers. The elements $s_1{s}_2$ and $s_2{s}_1$ are distinct of equal length $2$, so neither is below the other by the strict length increase in F2, and they are incomparable; likewise $s_2{s}_1\mathrm{̸}{\le }_R{s}_1$ and $s_1{s}_2\mathrm{̸}{\le }_R{s}_2$ by length, while $s_1{\le }_R{s}_2{s}_1$ would force $\mathrm{\ell }(s_2{s}_1)=\mathrm{\ell }(s_1)+\mathrm{\ell }(s_1^{-1}{s}_2{s}_1)=1+\mathrm{\ell }(w_0)=4$ by F2, which is false; so $s_1,s_2{s}_1$ are incomparable, and symmetrically $s_2,s_1{s}_2$ are incomparable. Consequently the subposet $\{1,s_1,s_1{s}_2,s_2{s}_1,w_0\}$ has the chains $1<s_1<s_1{s}_2<w_0$ and $1<s_2{s}_1<w_0$ together with the incomparabilities just listed, that is, it is the pentagon.

Failure of distributivity. Put $u=s_1$, $v=s_1{s}_2$ and $d=s_2{s}_1$. The upper bounds of $\{u,d\}$ are the elements above both: above $s_1$ lie $s_1,s_1{s}_2,w_0$ and above $s_2{s}_1$ lie $s_2{s}_1,w_0$, so the only common upper bound is $w_0$ and $u\vee d=w_0$. Since $v=s_1{s}_2{\le }_R{w}_0$, one has $v\wedge (u\vee d)=v\wedge w_0=v=s_1{s}_2$. The only reduced word of $v=s_1{s}_2$ is $(s_1,s_2)$: the only length-two words are $s_1{s}_1,s_1{s}_2,s_2{s}_1,s_2{s}_2$, the equal-letter words represent $1$, and $s_1{s}_2$ and $s_2{s}_1$ are distinct by the element list in 1.1. The only reduced word of $d=s_2{s}_1$ is likewise $(s_2,s_1)$. Therefore the elements below $v$ are $1,s_1,s_1{s}_2$, while those below $u=s_1$ are $1,s_1$, so $v\wedge u=s_1$; the elements below $d$ are $1,s_2,s_2{s}_1$, whose intersection with the elements below $v$ is just $1$, so $v\wedge d=1$. Hence $(v\wedge u)\vee (v\wedge d)=s_1\vee 1=s_1$, while $v\wedge (u\vee d)=s_1{s}_2\mathrm{\ne }{s}_1$; the distributive identity of [F6] fails for the triple $(u,v,d)$, so $[1,w_0{]}_R$ is not distributive.

By step 1.1, $s_1{s}_2{s}_1$ is a reduced word of $w_0$ containing the contiguous factor $⟨s_1,s_2{⟩}_3$, so by [F5] the element $w_0$ is not fully commutative; this exhibits a non-distributive right weak interval below a non-fully-commutative element. The interval theorem [F7] concerns only fully commutative elements, so no contradiction arises, and the example verifies only the failure of distributivity for this single interval; it does not prove the converse implication.