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.
Trouble Understanding a Remark about Beta-Reduction in the Lambda Calculus
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