Trouble Understanding a Remark about Beta-Reduction in the Lambda Calculus From page 15 of Lambda Calculus and Combinators an Introduction:


  Note 1.34 If $M \equiv aM_1 \ldots M_n$ where $a$ is an atom, and $M \triangleright_\beta N$, then $N$ must have form
  
  $$ N \equiv a N_1 \ldots N_n $$
  
  where $M_i \triangleright_\beta N_i$ for $i = 1, \ldots, n$. To see
  this, note that $M$ is really
  
  $$ (( \ldots ((aM_1)M_2) \ldots)M_n) $$
  
  when its parentheses are fully shown; hence each $\beta$-redex in $M$
  must be in $M_i$. Also the same holds for each subterm $\lambda x. P$
  whose bound variable might be changed in the reduction of$M$.


Question: Why is the following not a counter-example?

First let $M_1 = m_1 m_2 m_3$ for atomic constants $m_1$, $m_2$, $m_3$. Then let $N_1 = m_1$ and $N_2 = m_2 m_3$. Then:

$$
aM_1 \equiv a (m_1 m_2 m_3) \equiv (((am_1)m_2)m_3) \equiv ((a m_1)m_2 m_3) \triangleright_\beta a N_1 N_2
$$

Here, if I am correct, then we have a case where

$$
a M_1 \triangleright_\beta a N_1 N_2
$$

which seems to contradict the note.
Application in the lambda calculus is not associative. $m_1 m_2 m_3$ means $(m_1 m_2) m_3$ and is not equal to $m_1 (m_2 m_3)$.  So the second equality you state doesn't hold. $$a(m_1 m_2 m_3) \neq ((a m_1) m_2) m_3$$  Further $$a N_1 N_2 = (a m_1)(m_2 m_3) \neq (a m_1) m_2 m_3 = ((a m_1) m_2) m_3$$