A fixed point f of a fixed-point combinator would be a function that has itself as a fixed point: f(f) = f. The only such function I could come up with is id, which by definition has the apparently stronger property that id(x) = x for all x. Equivalently, everything is a fixed point of id.

My question is: is this actually a stronger property (in untyped lambda calculus), or is id the only function with f(f) = f?


1 Answer 1


If by "$=$" you mean $\beta$-equality, then the answer is yes, $MX=X$ for all $X$ is a stronger property than $MM=M$.

For example, let $$A := \lambda a.aa(aa)$$ (to save parentheses, I am using the standard left-associative notation for application; in your notation, the above term would be $\lambda a.a(a)(a(a))$) and take $$M := AA.$$ We clearly have $M\to MM$ and therefore $MM=M$. On the other hand, for any normal form $N$, $MN\neq N$, because $MN$ does not normalize (in fact, $M$ does not have a head normal form).

  • $\begingroup$ Thanks so much for the elegant counter-example. Do you know if the same would hold when only considering normalizing terms? I think it's an important restriction because a normalizing term is meaningful as a function and we could see how it differs semantically from the identity. $\endgroup$ Commented Jul 22, 2021 at 22:24
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    $\begingroup$ It's easy to see that $MM=M$ for $M$ normalizable implies that $NN\to^\ast N$ where $N$ is the normal form of $M$. I can't quickly come up with an example of normal term different from the identity which reduces to itself when applied to itself, but I'm not convinced it does not exist... I honestly don't know! $\endgroup$ Commented Jul 24, 2021 at 6:12

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