This page contains proofs to some assertions that are left as exercises in Combinator engine.
I = S K K
To prove this, remember that
S x y z = x z { y z }
and
K x y = x
So what is S K K p?
S K K p = K p { K p } = p
B = S {K S} K
Remember, we want
B x y z = x { y z }
So, let's see:
S { K S } K x y z = K S x { K x } y z = S { K x } y z = K x z { y z } = x { y z }
C = S { B B S } { K K }
We want to have
C f x y = f y x
({C f y} is a notation for currying the second argument of f, just as {f x} curries the first.)
S { B B S } { K K } f x y = B B S f { K K f } x y = B { S f } { K K f } x y = S f { K K f x } y = f y { K K f x y } = f y { K x y } = f y x