According to Peter Selinger, The Lambda Calculus is Algebraic (PDF). Early in this article he says:

The combinatory interpretation of the lambda calculus is known to be imperfect, because it does not satisfy the $ξ$-rule: under the interpretation, $M = N$ does not imply $\lambda x.M = \lambda x.N$ (Barendregt, 1984).


  • What kind of equivalence is meant here?
  • Given this definition of equivalence, what is a counter-example of the implication?

2 Answers 2


The equivalence is just equivalence in the equational $\lambda$-theory under discussion. In this case, it's the theory outlined in Table 1. Note that this theory does not include $\eta$: doing so would make the theory extensional, and the point is eventually that $\xi$ respects $\lambda$'s intensionality, while it would make CL partially extensional. I am not sure why the other answer mentions $\eta$.

Note that in $\lambda$:

$$(M =_\beta N) \Longrightarrow (\lambda x.M =_\beta \lambda x.N) \tag{1}$$

This should be intuitively obvious: if $M$ is $\beta$-convertible to $N$ when it stands by itself, then it is also $\beta$-convertible to $N$ when it is a subterm of $\lambda x.M$.

The $\xi$-rule, defined as \begin{align} M &\;= N \\ \hline (\lambda x.M) &\;= (\lambda x.N) \tag{$\xi_\lambda$} \end{align} makes this inference directly possible when it is part of a $\lambda$-theory. Its CL analogue would be: \begin{align} M &\;= N \\ \hline (\lambda^* x.M) &\;= (\lambda^* x.N) \tag{$\xi_{CL}$} \end{align}

Now, the point is that in CL, the following does not hold:

$$(M =_w N) \Longrightarrow (\lambda^* x.M =_w \lambda^*x.N) \tag{2}$$

In other words, if two terms are weakly equal, then this is not necessarily true for their pseudo-abstracted versions.

Consequently, if we add $\xi_{CL}$ to a CL theory, then we start equating terms which have different normal forms.

Note. Here, $M =_w N$ denotes weak equality. It means that $M$ can be converted into $N$ (and vice versa) by a series of $\mathsf{S}$ and $\mathsf{K}$ contractions (possibly also $\mathsf{I}$, if it is part of the theory). As you probably know, $=_w$ is the CL analogue of $=_\beta$.

$\lambda^*$ is the pseudo-abstractor as defined on page 5 of your document. It has the following property:

$$(\lambda^*x.M)N \rhd_w [N/x]M \tag{3}$$

This property makes it easy to find a CL analogue for any $\lambda$-term: just change $\lambda$ to $\lambda^*$ and apply the translations according to the definition of $\lambda^*$.

To be clear, the 'counter-example' in this answer is not a counter-example to (2). Because if we have:

$$M = x \tag{4}$$ $$N = (\lambda^*z.z)x \tag{5}$$

Then $N$ really denotes (applying the translations of page 5, and the fact that $\mathsf{I}$ is defined as $\mathsf{SKK}$ at the end of page 4):

$$N = (\lambda^*z.z)x = \mathsf{I}x = \mathsf{SKK}x \tag{6}$$

Since $\mathsf{SKK}x \rhd_w \mathsf{K}x(\mathsf{K}x) \rhd_w x$, we indeed have that $M =_w N$. However, if it is a counter-example, we should then have that $(\lambda^* y. M) \not=_w (\lambda^* y. N)$. But if we translate, we actually get:

$$(\lambda^* y. M) = (\lambda^* y. x) = \mathsf{K}x \tag{7}$$ $$(\lambda^* y. N) = (\lambda^* y. \mathsf{SKK}x) = \mathsf{K}(\mathsf{SKK}x)\tag{8}$$

And is easy to verify that (7) and (8) are still weakly equal, for:

$$\mathsf{K}(\mathsf{SKK}x) \rhd_w \mathsf{K}(\mathsf{K}x(\mathsf{K}x)) \rhd_w \mathsf{K}x \tag{9}$$

Now, a proper counter-example to (2) would be:

$$M = \mathsf{K}xy$$ $$N = x$$

Since $\mathsf{K}xy \rhd_w x$, we definitely have that $M =_w N$. However, if you translate carefully for the abstracted versions, then you will see that both are distinct normal forms - and these cannot be convertible according to the Church-Rosser theorem.

First we check $M'$:

\begin{align} M' &= \lambda^* x.\mathsf{K}xy \\ &= \mathsf{S}(\lambda^* x.\mathsf{K}x)(\lambda^* x.y)\\ &= \mathsf{S}(\lambda^* x.\mathsf{K}x)(\mathsf{K}y) \\ &= \mathsf{S}(\mathsf{S}(\lambda^* x.\mathsf{K})(\lambda^* x.x))(\mathsf{K}y) \\ &= \mathsf{S}(\mathsf{S}(\lambda^* x.\mathsf{K})(\mathsf{I}))(\mathsf{K}y) \\ &= \mathsf{S}(\mathsf{S}(\lambda^* x.\mathsf{K})(\mathsf{SKK}))(\mathsf{K}y) \\ &= \mathsf{S}(\mathsf{S}(\mathsf{KK})(\mathsf{SKK}))(\mathsf{K}y) \end{align} Here you can verify that $M'$ is a normal form. Here you can check that $(\lambda^* x.Kxy)P \rhd_w P$, as you should expect if $\lambda^*$ is supposed to behave like an abstractor for CL. (Click the blue links to perform weak contractions.)

Now we check $N'$: \begin{align} N' &= \lambda^*x.x \\ &= \mathsf{I} \\ &= \mathsf{SKK} \end{align}

Which is obviously a normal form different from $M'$, so $M' \not=_w N'$ by the Church-Rosser theorem. Note also that $N'P \rhd_w P$, i.e. $M'$ and $N'$ 'produce the same output' for arbitrary inputs $P$.

We have now proven that (2) does not hold in CL, and that a CL theory incorporating $\xi$ would therefore equate terms that are not weakly equal. But why do we care?

Well, first of all, it makes the combinatory interpretation of $\lambda$ imperfect: apparently not all metatheoretic properties carry over.

In addition, and perhaps more importantly, while there exist extensional theories of $\lambda$ and CL, they are originally and commonly kept intensional. Intensionality is a nice property because $\lambda$ and CL model computation as process, and from this perspective two different programs (specifically, terms that have a different normal form) that always produce the same results (given equal inputs) are not to be equated. $\xi$ respects this principle in $\lambda$, and if we want to make $\lambda$ extensional, we could just add e.g. $\eta$. But the introduction of $\xi$ in CL would no longer make it completely intensional (in fact, only partially so). And this is the reason for $\xi$'s 'notoriety', as the article puts it.

  • 1
    $\begingroup$ I can't comment on quality because I know little of the subject, but this looks like a bit of work. Appreciated, thanks! $\endgroup$
    – Raphael
    Jul 23, 2014 at 17:20
  • $\begingroup$ Indeed, the post ended up longer than I had anticipated. Thanks for your comment. :) $\endgroup$
    – Roy
    Jul 23, 2014 at 18:12
  • 2
    $\begingroup$ Oh, that. Happens. Regularly. $\endgroup$
    – Raphael
    Jul 23, 2014 at 18:24

EDIT This answer is incorrect, as the other answerer correctly pointed out. I used the translation into combinatory logic from Asperti & Longo, which is subtly different from the one in Selinger.

In fact, this illustrates a crucial point: "the combinatory interpretation" of lambda calculus is not a single thing! Different authors do it slightly differently.

I'm leaving my answer here for posterity, but the other answer is better.

Equivalence in this context is defined by Tables 1 and 2 in Selinger's paper. However, a slightly different axiomatisation may make things a little more clear.

What it really means is that two terms are convertible in the $\lambda$ theory. We can define "convertibility" by the following two axioms:

  • $\beta$. $(\lambda x. M) N = [N/x]M$, if $x$ free for $N$ in $M$
  • $\eta$. $\lambda y. M y = M$, if $y$ not free in $M$

plus, of course, the usual axioms and inference rules needed to make $=$ a congruence. From this, it should be obvious that any counter-example is going to rely on the free variable condition on the $\eta$ rule being broken.

I think this is probably the simplest:

$$M = x$$ $$N = (\lambda z. z) x$$

You can verify for yourself that $\lambda y. M = \lambda y. N$, but their respective combinatorial interpretations are not equal under the rules in Table 2.

  • $\begingroup$ What I don't understand about your answer: 1) why mention $\eta$, while the theory in Table 1 does not include it and is clearly intensional? 2) How are the combinatory interpretations of $\lambda y.M$ and $\lambda y. N$ not equal? The derivation in my answer shows that they are. 3) The $\xi$-rule is not addressed, while that is the culprit in the issue. $\endgroup$
    – Roy
    Jul 24, 2014 at 0:17

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