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In Sorensen and Urzyczyn's book there is a lemma (1.3.11) which I am having a hard time proving.

1.3.11 Lemma:

If there is an infinite $\beta \eta$-reduction sequence starting with a term $M$ then there is an infinite $\beta$-reduction seqeunce from $M$.

Now the proof is sketched in the book but I am having trouble realising it:

Proof. First observe that in an infinite $\beta\eta$-reduction sequence there must be infinitely many $\beta$-reduction steps (cf. Exercise 1.6). These $\beta$-reduction steps can be "permuted forward", yielding an infinite $\beta$-reduction. Indeed, by induction with respect to $M \to_{\eta} N$, one can show that $M \to_{\eta}N \to_{\beta}L$ implies $M \to_\beta P \twoheadrightarrow_{\beta \eta} L$, for some $P$.

So $\to_\eta$ is defined as the least compatible relation satisfying $\lambda x . f x \to_\eta f$. Where compatible means works with the syntax, i.e. $A \to B$ imples $M A \to M B$, $A M \to B M$ and $\lambda x . A \to \lambda x .B$.

The definition is similar for $\beta$ with the obvious reduction. The relation $\to_{\beta \eta}$ is the union of both. And $\twoheadrightarrow$ is the transitive-relfexive closure of said relation.

Exercise 1.6 was to prove that $\to_\eta$ is strongly normalising, which can be shown by considering a "depth" function on terms, and showing that this strictly decreases on $\eta$-reduction.

The problem I am having is I can't seem to understand how $\beta$ and $\eta$ interact.

Take for example $\lambda t. (\lambda x . y) t$. Clearly we can $\beta$-reduce this to get $\lambda t . y$ or $\eta$-reduce it to get $\lambda x . y$ which are clearly equal but its not obvious to me in general.

How can this be proven? And better yet, are there any references for this? I can't seem to find much on $\eta$-reduction in lambda calculus.

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  • $\begingroup$ Your example $\lambda t. (\lambda x . y) t$ does not $\eta$-reduce to $\lambda x . y$, provided $f$, in your rendition $\lambda x . f x \to_\eta f$ of the $\eta$-axiom, ranges over variables rather than terms. $\endgroup$
    – Martin Berger
    Apr 27, 2019 at 7:11
  • $\begingroup$ @MartinBerger If you restrict $\eta$-reduction to variables, how do you reduce $\lambda x.(yz)x$ to the $\beta\eta$-equivalent term $(yz)$? $\endgroup$
    – Emil Jeřábek
    Apr 27, 2019 at 8:40
  • $\begingroup$ Regarding the orginal question: the key insight is that $\eta$-reduction never removes any (interesting) $\beta$-redexes, so any $\beta$-redex that exists in term $N$ after an $\eta$-reduction $M \rightarrow_{\eta} N$, existed already in $M$. Hence $M$ has a $\beta$-reduction. Then you appeal to something like Church-Rosser. $\endgroup$
    – Martin Berger
    Apr 27, 2019 at 10:26
  • $\begingroup$ @EmilJeřábek Note that the standard presentation of the $\eta$ axiom (e.g. in Barendregt's The Lambda Calculus, or Hindley/Seldin's Lambda-Calculus and Combinators) is: $\lambda x.Mx =_{\eta} M$ provided $x$ isn't free in $M$. The two forms of $\eta$ are equivalent when considered as axioms for an equational theory. Why? Because clearly: $(\lambda x.Mx) =_{\beta} (\lambda fx.fx)M$ $=_{\eta} (\lambda f.f)M =_{\beta} M$. In this derivation, the second step uses $=_{\eta}$ with variables only. The side-condition that $x$ not be free in $M$ is used in the first step. $\endgroup$
    – Martin Berger
    Apr 27, 2019 at 10:35
  • $\begingroup$ @MartinBerger I know all that. However, the whole point of reductions (as opposed to equations) is that you can apply them only in one direction. The $\beta\eta$-reduction calculus will fail to have the Church–Rosser property and fail to be equivalent to the $\beta\eta$-equational theory if you restrict the $\eta$-rule to variables. $\endgroup$
    – Emil Jeřábek
    Apr 27, 2019 at 11:10

1 Answer 1

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As usual, the bible knows all, and Lemma 3.3.8 (p 66) gives the detailed proof of commutativity of $\beta$ and $\eta$ reductions.

Trying to reproduce it here would be tedious and wouldn't do it justice, so I'll include a picture:

enter image description here

You had the main idea right, it just takes a bit of practice to generalize from the specific examples.

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