Can XOR be expressed using SKI combinators?

Question

I have question about SKI-Combinators.

Can XOR (exclusive or) be expressed using S and K combinators only?

I have

True = Cancel
False = (Swap Cancel)

where

Cancel x y = K x y = x   
Swap: ff x y = S ff x y = ff y x

Answer 1

Booleans

Your question is a bit unclear on the details, but it seems that what you mean is that you have the following representation of booleans:

T := K
F := S K

This works because it means the following reductions hold:

T t e => t
F t e => e

in other words, b t e can be interpreted as IF b THEN t ELSE e.

XOR in terms of IF _ THEN _ ELSE _

So given this framework, how do we implement XOR? We can formulate XOR as an IF expression:

xor x y := IF x THEN (not y) ELSE y = (IF x THEN not ELSE id) y

which can be eta-reduced to

XOR x := IF x THEN not ELSE id = x not id

Some function combinators

We have id = SKK as standard, and not can be expressed as flip, since flip b t e = b e t = IF b THEN e ELSE t = IF (not b) THEN t ELSE e. flip it self is quite involved but doable as

flip := S (S (K (S (KS) K)) S) (KK)

Now we just need to figure out a way to write a function that takes x and applies it on the two terms NOT and ID. To get there, we first note that if we set

app := id

then

app f x = (id f) x = f x

and so,

(flip app) x f = f x

We are almost there, since everything so far shows that

((flip app) id) ((flip app) not x) = ((flip app) not x) id = (x not) id = x not id

The last step is to make that last line point-free on x. We can do that with a function composition operator:

((flip app) id) ((flip app) not x) = compose ((flip app) id) ((flip app) not) x

where the requirement on compose is that

compose f g x = f (g x)

which we can get by setting

compose f g := S (K f) g

Putting it all together

To summarize, we got

xor := compose ((flip app) id) ((flip app) not)

or, fully expanded:

xor = S (K ((flip app) id)) ((flip app) not)
    = S (K ((flip app) (SKK))) ((flip app) flip)
    = S (K ((flip SKK) (SKK))) ((flip SKK) flip)
    = S (K (((S (S (K (S (KS) K)) S) (KK)) SKK) (SKK))) (((S (S (K (S (KS) K)) S) (KK)) SKK) (S (S (K (S (KS) K)) S) (KK)))

Answer 2

Yes! In 2020, Stephen Wolfram constructed combinators for all 16 two-input Boolean functions, including XOR. It turns out XOR can be expressed as s[s[s[s[s]][s[s[s[k]]]][s]]][k] where k represents the value True and s[k] represents the value False.

For example, here is the combinator form of XOR operating on true and false the Wolfram Language.

true = k;
false = s[k];
ResourceFunction["CombinatorFixedPoint"][s[s[s[s[s]][s[s[s[k]]]][s]]][k][true][false], SKGlyphs -> {s, k}]

It returns the expected output – True:

k