SKI Combinator Calculus - Informal Description

Informal Description

Informally, and using programming language jargon, a tree (xy) can be thought of as a "function" x applied to an "argument" y. When "evaluated" (i.e., when the function is "applied" to the argument), the tree "returns a value", i.e., transforms into another tree. Of course, all three of the "function", the "argument" and the "value" are either combinators, or binary trees, and if they are binary trees they too may be thought of as functions whenever the need arises.

The evaluation operation is defined as follows:

(x, y, and z represent expressions made from the functions S, K, and I, and set values):

I returns its argument:

Ix = x

K, when applied to any argument x, yields a one-argument constant function Kx, which, when applied to any argument, returns x:

Kxy = x

S is a substitution operator. It takes three arguments and then returns the first argument applied to the third, which is then applied to the result of the second argument applied to the third. More clearly:

Sxyz = xz(yz)

Example computation: SKSK evaluates to KK(SK) by the S-rule. Then if we evaluate KK(SK), we get K by the K-rule. As no further rule can be applied, the computation halts here.

Note that, for all trees x and all trees y, SKxy will always evaluate to y in two steps, Ky(xy) = y, so the ultimate result of evaluating SKxy will always equal the result of evaluating y. We say that SKx and I are "functionally equivalent" because they always yield the same result when applied to any y.

Note that from these definitions it can be shown that SKI calculus is not the minimum system that can fully perform the computations of lambda calculus, as all occurrences of I in any expression can be replaced by (SKK) or (SKS) or (SK whatever) and the resulting expression will yield the same result. So the "I" is merely syntactic sugar.

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:

ιx = xSK