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.

  • $\begingroup$ I think you are missing an $a$ somewhere in the first statement. $\endgroup$ Jan 14, 2017 at 16:29
  • $\begingroup$ @YuvalFilmus: You're correct. I just fixed it. $\endgroup$
    – George
    Jan 14, 2017 at 16:30

1 Answer 1


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 $$a N_1 N_2 = (a m_1)(m_2 m_3) \neq (a m_1) m_2 m_3 = ((a m_1) m_2) m_3$$


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