# Is there a right identity for Application in Lambda calculus?

Such a function E that:

∀F (F E = F)


It's obviously, that the left identity E' (E' F = F) is a function (λx.x). Indeed: (λx.x) F == F.

But is there a right one like that?

Or is there evidence that such an element does not exist?

In common case, is there a chain of elements (E1 E2 … En) for every well-formed expression (function) of Lambda calculus which fulfills the condition:

∀F (F E1 E2 … En = F) (n >=1)


Is there a method to build such a chain?

PS For some functions it is easy, for example:

Let F := (λa.λb.a b), then E1 := (λс.c) and E2 := F:
(λa.λb.a b) (λс.c) (λa.λb.a b) =>β
(λb.(λс.c) b) (λa.λb.a b) =>β
(λb.b) (λa.λb.a b) =>α
(λb.b) (λa.λc.a c) =>β
λa.λc.a c) =>α
(λa.λb.a b)


but what about, for example (λa.λb.a b a)?

If I understand correctly, the answer is definitely no - at least, in any reasonable context.

The short version: think about any constant map $$\lambda x.p$$, and what the existence of a right identity says about this map.

In more detail:

For simplicity I'll play a bit fast and loose with the details below.

I'll take as my context an arbitrary partial applicative structure $$(A,\cdot)$$. First, suppose our PAS is total. A reasonable requirement is for constant maps to exist: for each $$a\in A$$ there should be some $$c_a\in A$$ with $$c_ax=a$$ for all $$x\in A$$. We now have a problem. If each element $$x$$ of $$A$$ has a right identity $$e_x$$, then we must have $$c_ae_{c_a}=c_a$$ for each $$a\in A$$. But $$c_ae_{c_a}=a$$ for each $$a\in A$$, so we get $$a=c_a$$ for every $$a\in A$$. We are thus in the degenerate situation where every functional is constant.

So total PASs don't have right identities unless they're extremely trivial, and in particular there is exactly one (fine, up to isomorphism) PAS of a given cardinality which has right identities. But essentially the same problem happens even if we drop totality. Call an element $$a\in A$$ almost constant if there is some $$v_a$$ such that $$ab=v_a$$ whenever $$ab$$ is defined. Then - yet again - a right fixed point $$c$$ for $$a$$ must make $$ac$$ defined, and so we get $$ac=v_a=a$$. In particular, for each $$v$$ there is at most one functional in $$A$$ which is only ever $$v$$ when defined, and that element is $$v$$ itself. At this point the mere existence of a "nearly constant" map for an arbitrary $$v\in A$$ collapses everything yet again: an element $$a$$ of $$A$$ is characterized completely by the subset $$D_a$$ of $$A$$ on which it is defined, and for each $$d\in D_a$$ we have $$ad=a$$. This does give more variety than the total case, but not much more, and in particular I think any reasonable semantics for anything like $$\lambda$$-calculus will not be satisfied by such a thing.

And in particular, no nontrivial partial combinatory algebra - and these are to my understanding the "best" PASs - can have this property. First, note that for each $$a$$ the object $$Ka$$ is equal to $$a$$ on all inputs (that is, $$Kab=a$$ for all $$b$$); by the above, this means $$Ka=a$$. In particular, since $$Ka$$ is total we get that each $$a\in A$$ behaves as the constant function $$\lambda x.a$$. In particular, our algebra is total. But now consider the $$S$$-object. By the above we have $$Sa=S$$ for every $$a$$. However, by the above we also know that $$(ac)(bc)=ab=a$$ for all $$a,b,c\in A$$; the $$S$$-axiom then tells us that $$Sabc=(ac)(bc)=a$$ for all $$a,b,c\in A$$. But $$Sabc=((Sa)b)c=S$$, and so we get $$a=S$$ - that is, our PCA has only one element.

• @Noah_Schweber Thank you very much ! – asianirish Dec 18 '18 at 14:14