Probably the most famous example of a divergent term (ie, one which admits infinitely many $\beta$-reductions) in the $\lambda$-calculus is the Y combinator

$$ Y = \lambda f. (\lambda x. f(xx)) (\lambda x. f(xx)) $$

The kind of the "heart" of the recursion here is the inclusion of the subterm $xx$. I wondered if this is true in general; ie, if we have the theorem

$$ E \text{ is a divergent $\lambda$-term} \implies E \text{ contains a subterm of the form } AA $$

(Note: $A$ can be any term, not just names as in $A = x$).

The answer is no. The following term reduces to itself (and thus diverges), but does not contain a subterm of the form $AA$.

$$ \begin{align*} &(\lambda ab.bab) (\lambda ab.aba) (\lambda ab.bab) \\ \rightarrow_\beta& (\lambda b.b(\lambda ab.aba)b)(\lambda ab.bab) \\ \rightarrow_\beta& (\lambda ab.bab) (\lambda ab.aba) (\lambda ab.bab) \end{align*} $$

However, this term does contain the subterm $bab$, which although not of the form $AA$ is of the form $AYA$, which also looks suspicious.

Is there any such "syntactic" condition on divergent terms?

Here's a conjecture: if $E$ is a divergent $\lambda$-term, then $E$ contains a subterm $S$ of the form $S = F A_1 A_2 \cdots A_n$ where at least one $A_i$ contains $F$ as a subterm. That is, "for a term to be divergent, it must afford itself the possibility for a subterm $F$ to receive itself as an argument"

(The converse is false, as $(\lambda x.xx)(\lambda a.a)$ satisfies this via inclusion of $xx$ but does not diverge)


1 Answer 1


This is essentially the criterion used by type systems to avoid divergence. Take, for instance, the simply typed lambda calculus. In order for $f\ x$ to be well typed, we must have:

$$f : T → U \\ x : T$$

In order for $f\ f$ to be well typed, it would have to be the case that:

$$T = T → U$$

But types are required to be inductively generated by $→$ and some base types, so there is no type that satisfies this equation. Similarly, in order for $f$ to receive itself as an argument at all, it would have to be the case that:

$$T = \cdots → T → \cdots$$

which is not allowed.

More elaborate type systems can allow some cases of self-application without enabling divergence. For instance, in the polymorphic lambda calculus, we can write:

$$λ(f : ∀T. T → T). f_{∀T. T → T}\ f$$

In this case, it is safe because $f$ can only be the identity function. Or, even more precisely, intersection types allow cases of self-application like so:

$$λ(f : T \cap (T → U)). f f$$

and these sort of intersection types classify exactly the normalizing lambda terms. I suspect the intuition is that for any normalizing term involving self-application, only finitely many unfoldings of the $T = \cdots → T → \cdots$ equation are necessary, and can be expressed by a finite intersection, while the diverging terms genuinely require an infinite type that cannot be sufficiently approximated by any finite intersection.

  • $\begingroup$ This is great, thank you! $\endgroup$
    – Quelklef
    Jun 22, 2022 at 16:08

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