Does SKS equal SKK?

Question

Context

I started teaching myself lambda calculus last night and I am trying to determine if what I understand so far is correct.

Understanding

SKK is equivalent to the Identity combinator, I.

Where L stands for lambda:

S = LxLyLz((xz)(yz))

K = LxLy(x)

K essentially takes the next 2 (lambda) terms and gives back the first of those. S seems a little more complicated in the untyped lambda calculus.

My Interpretation

SK(any-lambda-term) is also equivalent to I.

I.e. the application of the application of S to K to Any-lambda-term is equivalent to the Identity combinator:

((S K)(Any)) = I = S K K = ((S K)(K))

I am using the convention of “left-association” in my above notation, if that helps (And I tried to make that clear in the 4th term above with parentheses. Everything I have read so far seems to use this convention).

Reasoning

S K = LyLz((K z)(y z))

The next lambda term will be substituted for y, let the term be Y.

S K Y = Lz((K z)(Y z))

(Y z) is the application of Y to z, also a lambda term. (K z)returns the constant function that returns z, given another term input: (Y z).

Is my interpretation true? If not, can you provide an explanation? I would greatly appreciate it. Particularly if a sort of order of operations can be explained—I regularly find myself confused when considering when to evaluate. Perhaps that will be refined with practice.

Answer 1

Your intuition is correct, but an intuition proves nothing (alas...)

So, how can we prove your statement? Simply by showing that SKK and SKS have the same behaviour. "Behaviour" is an informal notion, which is formally capture by "semantics": if SKK and SKS are equals, then they should always reduce to the same term, according to the SKI-calculus semantics.

Now, there is a deep question, which is: what are the SKI-calculus? Actually, there is not a single way to answer that. What you implicitly do in your question is that you express SKI in terms of λ terms and you rely on the semantics of the λ calculus. This is absolutly correct. An other way to do it could have been to define directly SKI semantics. For instance, if you look at the wikipedia page, you can see that the semantics are not defined with lambda terms (and the fact that it correspond to lambda term is a (nice and expected) side effect). In the rest of this answer, I'll take the same approach as you do, and convert SKI terms in λ terms. A good exercise for you is to redo the proof, using the proper SKI semantics.

So, let formalize your question: your question is whether, for any SKI term t, SKKt = SKSt? Well... Let's see.

SKKt is encoded as (λx.λy.λz.(xz)(yz))(λx.λy.x)(λx.λy.x)t in the λ-calculus. We now just have to reduce it to a normal form (I detail every step, each time I reduce the leftmost λ, even tho it is not the fastest strategy):

(λx.λy.λz.(xz)(yz))(λx.λy.x)(λx.λy.x)t
= (λy.λz.((λx.λy.x)z)(yz))(λx.λy.x)t
= (λz.((λx.λy.x)z)((λx.λy.x)z))t
= ((λx.λy.x)t)((λx.λy.x)t)
= (λy.t)((λx.λy.x)t)
= t

So, the encoding of SKKt in the λ calculus reduces to t (as a sidenote, we just proved that SKK is equivalent to I here). To conclude our proof, we have to reduce SKSt and see whether it also reduces to t.

SKSt is encoded as (λx.λy.λz.(xz)(yz))(λx.λy.x)(λx.λy.λz.(xz)(yz))t. Let reduce it. (I don't detail as much this time)

(λx.λy.λz.(xz)(yz))(λx.λy.x)(λx.λy.λz.(xz)(yz))t
= ((λx.λy.x) t)((λx.λy.λz.(xz)(yz)) t))
= (λy.t)((λx.λy.λz.(xz)(yz)) t))
= t

Hurrah! It also reduce to t, so indeed, SKS and SKK are equivalent. It seems that the third combinator is not important: that as soon as you have SK?, it is equivalent to I. As an exercise, you can easily prove it (same strategy, if it is the case, then for any terms t and s, SKts = s). As mentionned above, an other good exercise is to redo the proof without using the λ semantics, but the proper SKI semantics.

Finally, my answer should raise a new question to you: we have two semantics, one that encodes SKI terms into λ terms, and one that does not. The question you may have is: are the two semantics equivalent? What does it mean for two semantics to be equivalent? If you are only starting to teach yourself λ calculus, it may be a bit early to try to answer those questions right now, but you can keep it in a corner of your head for when you'll get more familiar with formal languages.