Let M be a von Neumann algebra with faithful normal normalized trace tr. Let $\{ e_i | i=1,2,\dots \}$ be projections in M such that:
$e_ie_{i \pm 1}e_i=\tau e_i $ for some $\tau \leq 1$
$e_ie_j=e_je_i $ for $|i-j| \geq 2$
$\text{tr}(we_i)=\tau \text{tr}(w)$ if $w$ is a word on $1,e_1,\dots,e_{i-1}$.
Let $A_n$ be the algebra generated by $1,e_1,\dots,e_n$ and $s_n=e_1\vee e_2 \dots \vee e_n$. I need to prove that if $s_n \neq 1$ then $e_{n+1} \wedge s_n = e_{n+1}s_{n-1}$.
Further details can be found at page 16 (4.2.2) of http://gdz.sub.uni-goettingen.de/dms/load/img/?PID=GDZPPN002099268
I tried to read the proof given in the paper but I couldn't understand it. Thank you for your time and help.
I can't blame you, the proof is written horribly (let's hope Vaughan doesn't read me on this...)
First you have, from 4.1.3, that $e_{n+1}s_ne_{n+1}=E_{A_{n-1}}(s_n)e_{n+1}$. From 4.2.1 you have $ws_n=w$ for all $w\in A_{n-1}$, and so you deduce that $E_{A_{n-1}}(s_n)\in Z(A_{n-1})$. Also from 4.2.1 you know that $1-s_{n-1}$ is a minimal projection in $Z(A_{n-1})$ (which is finite-dimensional by 4.1.3), so we can write $$ E_{A_{n-1}}(s_n)=\lambda_0(1-s_{n-1})+\sum_{j=1}^k\lambda_jp_j, $$ with $p_1,\ldots,p_k$ minimal projections in $Z(A_{n-1})$ with $p_1+\cdots+p_k=s_{n-1}$, and $\lambda_0,\ldots,\lambda_k\in[0,1]$.
From the usual formula $p\wedge q=\lim_m(pqp)^m$ we get \begin{align} e_{n+1}\wedge s_n&=\lim_m(e_{n+1}s_ne_{n+1})^m =e_{n+1}\,\lim_m\left(\lambda_0(1-s_{n-1})+\sum_{j=1}^k\lambda_jp_j\right)^m\\ &=e_{n+1}\,\lim_m\left(\lambda_0^m(1-s_{n-1})+\sum_{j=1}^k\lambda_j^mp_j\right)\\ &=e_{n+1}\,\left(\delta_0(1-s_{n-1})+\sum_{j=1}^k\delta_jp_j\right),\\ \end{align} where $\delta_j=\lim_m\lambda_j^m\in\{0,1\}$. We have the obvious inequality $e_{n+1}\wedge s_n\geq e_{n+1}s_{n-1}$; as multiplication by $e_{n+1}$ is an isomorphism on $A_{n-1}$ (by (iii) in 4.1.3), we obtain $$ \delta_0(1-s_{n-1})+\sum_{j=1}^k\delta_jp_j\geq s_{n-1}=\sum_{j=1}^kp_j. $$ We deduce that $\delta_1=\cdots=\delta_k=1$, which in turn implies $\lambda_1=\cdots=\lambda_k=1$. Thus $$ E_{A_{n-1}}(s_n)=\lambda_0(1-s_{n-1})+s_{n-1}. $$ The hypothesis $s_n\ne1$ forces $E_{A_{n-1}}(s_n)\ne1$ (because the expectation is faithful). This means that $\lambda_0<1$, and so $\delta_0=0$. So $$ e_{n+1}\wedge s_n=e_{n+1}\sum_{j=1}^np_j=e_{n+1}s_{n-1}. $$