Comments on: S Combinator is Injective, with Proofs

By: -= Linkage 2007.12.13 =-

-= Linkage 2007.12.13 =- — Mon, 26 Jan 2009 15:40:45 +0000

[...] S Combinator is injective
Mathematics, Mathematica and Certainty
Pure functional data structures
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By: Shin

Shin — Sun, 28 Sep 2008 16:10:08 +0000

Loren S, thanks for pointing that out!

By: Loren S

Loren S — Mon, 25 Aug 2008 14:04:56 +0000

As a general matter, a left inverse of a (necessarily injective) function is always also a right inverse on the range of the function. That is, if g f = Id, then f g(t) = f g f(s) = f(s) = t for t = f(s) in the range of f. This makes your demonstration just before “Inverting higher order functions” unnecessary.

By: Charles Stewart

Charles Stewart — Tue, 11 Dec 2007 12:17:59 +0000

I've posted this observation as a story on LtU.

By: Shin

Shin — Thu, 06 Dec 2007 01:50:51 +0000

lf, Yes, I know about Djinn! In fact when I thought I found a general way to invert functions, I was merely constructing a term having the right type. After my talk, one of the audience fired up Djinn and showed me how it can be done. It must be one of my worst talks.. Frank Atanassow, Oh, I've probably omitted too many steps. The reasoning goes like this:


   (∀ y ∈ A -> B : P)
≡  (∀ y : y ∈ A -> B ⇒ P)
⇒   { since (∃ b : y = K b) ⇒ (y ∈ A -> B) }
   (∀ y : (∃ b : y = K b) ⇒ P)
≡    { since (∃ b : Q) ⇒ P ≡ (∀ b : Q ⇒ P) }
   (∀ y : (∀ b : y = K b ⇒ P))
≡  (∀ b : P[(K b)/y])

Chris Rathman, Thanks! I will have a read. The team I used to work with was also working on bidirectionality and I have to read that paper anyway...

By: Chris Rathman

Chris Rathman — Wed, 05 Dec 2007 15:10:02 +0000

Off hand, Benjamin Pierce’s work on Bijection and Bidirectional programming come to mind. See: http://lambda-the-ultimate.org/node/1526

By: Frank Atanassow

Frank Atanassow — Wed, 05 Dec 2007 13:49:17 +0000

In your first proof, it looks to me like b should be existentially quantified. Why do you say it is universally quantified?

By: lf

lf — Wed, 05 Dec 2007 11:43:58 +0000

Try Djinn! Passing it the type of S^-1:

f ? ((a->b) -> (a->c)) -> (a -> b -> c)

it gives you

f a b c = a (\ _ -> c) b

right away :-)