Talk:SKI combinator calculus

In fact, it is possible to define a complete system using only one combinator. An example is Chris Barker's iota combinator, defined as follows:

ix=x\mathbf {SK}

I find this unconvincing - $i$ is defined in terms of S and K, which means the system needs three combinators. If you could define i in terms of itself only, then I'd be convinced. I doubt this is possible, but I would be pleased if someone could prove me wrong! The iota language in the link uses S and K internally, it's just in the syntax that it is forbidden.

So I think this might mislead readers into thinking that single combinator systems are possible. Can anyone provide a more compelling example or clarify somehow?

Edwin 20:17, 14 May 2006 (UTC)[reply]

JA: The statement above is just the usual reduction of I to S and K, which gets it down to two. The reduction to one is rather artificial, but does it in terms of a combinator J. You can find that discussed in van Heijenoort's anthology. Jon Awbrey 20:40, 14 May 2006 (UTC)[reply]

Not

I'm probably missing something, but i'm confused by the postfix NOT in the article. it says "(T) NOT = T(F)(T) = F", but NOT is a unary function that applies its (binary function) argument to T and F, and T is a binary function that returns its first argument, so what we really want to do is apply NOT to T. Am I getting confused by the notation? Example in ML:

fun T x y = x;
fun F x y = y;
fun NOT x = x F T;

(NOT T) 1 2 -> 2 (it returned the second argument, so it's F)

Moe Aboulkheir 07:21, 18 August 2006 (UTC)[reply]