I understand that a Church numeral $c_n$ looks like $\lambda s. \lambda z. s$ (... n times ...) $s\;z$. This means nothing more than "the function $s$ applied $n$ times to the function $z$".

A possible definition of the $\mathtt{times}$ function is the following: $\mathtt{times} = \lambda m. \lambda n. \lambda s. m \; (n\; s)$. Looking at the body, I understand the logic behind the function. However, when I start evaluating, I get stuck. I will illustrate it with an example:

$$\begin{align*} (\lambda m. \lambda n. \lambda s. m \; (n\; s))(\lambda s.\lambda z.s\;s\;z)(\lambda s.\lambda z.s\;s\;s\;z) \mspace{-4em} \\ \to^*& \lambda s. (\lambda s.\lambda z.s\;s\;z) \; ((\lambda s.\lambda z.s\;s\;s\;z)\; s)) \\ \to^*& \lambda s. (\lambda s.\lambda z.s\;s\;z) \; (\lambda z.s\;s\;s\;z) \\ \to^*& \lambda s. \lambda z.(\lambda z.s\;s\;s\;z)\;(\lambda z.s\;s\;s\;z)\;z \end{align*}$$

Now in this situation, if I first apply $(\lambda z.s\;s\;s\;z)\;z$, I get to the desired result. However, if I apply $(\lambda z.s\;s\;s\;z)\;(\lambda z.s\;s\;s\;z)$ first, as I should because application is associative from the left, I get a wrong result:

$\lambda s. \lambda z.(\lambda z.s\;s\;s\;z)\;(\lambda z.s\;s\;s\;z)\;z \to \lambda s. \lambda z.(s\;s\;s\;(\lambda z.s\;s\;s\;z))\;\;z$

I can no longer reduce this. What am I doing wrong? The result should be $\lambda s. \lambda z.s\;s\;s\;s\;s\;s\;z$

  • $\begingroup$ The Church numerals in your starting term aren't right. 2 is represented by $\lambda s. \lambda z. s (s z)$, not $\lambda s. \lambda z. s s z$. $\endgroup$
    – Uday Reddy
    Apr 15, 2012 at 20:29

1 Answer 1


I think your reduction is correct (I've only eyeballed it, though). At the end, you can't apply $(\lambda z. s s s z)$ to $z$, this never appears in the term. $\lambda z. f f z$ is $\lambda z. (f f) z$, not $\lambda z. f (f z)$. Functions in the lambda-calculus take a single argument; they are effectively curried: a two-argument function is implemented as a one-argument function that takes the first argument and returns a new one-argument function that takes the second argument and returns the result.

You made the same mistake when defining Church numerals. The Church numeral for $n$ is based on composing a function $n$ times. “The function $s$ applied $n$ times to the function $z$” $\lambda s. \lambda z. s (s ( … s \: z) … ))$. What you wrote is the function $s$ applied $n-1$ times to the function $s$ and finally to $z$, which doesn't strike me as a useful term.

$2 \times 3$ is thus $(\lambda m n s. m (n s)) (\lambda s z. s (s \: z)) (\lambda s z. s (s (s \: z)))$. I'll let you check that it does reduce to $\lambda s z. s (s (s (s (s (s \: z)))))$.

  • $\begingroup$ As far as your paragraph goes, you are right and I was aware of this. It just striked me that applying rightly associative did yield the right result. As far as the second paragraph: you are right. Using no braces was erroneous of me, again because of the application's left associativity. I will reduce the whole thing again now and see if my lack of braces did cause my error! $\endgroup$
    – codd
    Apr 14, 2012 at 0:37
  • $\begingroup$ It did. You noting that my notation implied wrong order of application solved the problem! Accepting your answer. $\endgroup$
    – codd
    Apr 14, 2012 at 0:51

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