The way I see it, it should not be further reducible. I'm thinking λz.zq is like

lambda z: z(q) # Python, not lambda calc

and you do not know z nor q so it is not further reducible.

The below derivation (where I am wondering about the last steps) is from the new Haskell book:

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1 Answer 1


It is $\beta$-reduction defined in terms of substitution (rename ($\alpha$-conversion) first if necessary):

$$(\lambda z. zq) (q) \to_{\beta} [z/q](zq) = qq. $$

In Haskell, $\lambda z.zq$ defines an anonymous function which takes a single parameter (represented by $z$) and performes $zq$.

  • $\begingroup$ Now that I understand the mechanics of lambda calc better I see that this was just a straightforward application of the rules. Thanks though. $\endgroup$ Jan 30, 2016 at 11:49

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