Defining the Y combinator in terms of S, K and I

Question

We know that the Y-combinator is defined as: $$\text{Y}:=\lambda f.(\lambda x.f(xx))(\lambda x.f(xx))$$

Wikipedia says :$$\text{Y}:=\text{S(K(SII))(S(S(KS)K)(K(SII)))}$$

Now the question is: What logical steps can we take to convert the first definition to the second?

While it is easy to show the equivalence between the two definitions, finding how the first definition can motivate and lead to the second definition is, in my opinion, a tricky task. I have added my proof as an answer, but all other ideas and suggestions are welcome.

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Note: This question was marked as duplicate of another. However, in my opinion, that question discusses the intuition behind the $\lambda$-calculus definition of the Y-combinator, and nowhere does it discuss about the combinatorial representation. My question, on the other hand, asks specifically about the conversion from the $\lambda$-representation to the combinatorial representation.

I believe these to be two very different questions, since the first one is “restricted” to the Y-combinator, while my question can (in essence) be generalized to proving equivalence between any lambda term and its respective combinatorial form.

Answer 1

Let's define $$\text{E}=\lambda\text{x. f (x x)}$$ which leads to:

$$ \begin{align*} \text{E x}&=\text{f (x x)}\\ &=\text{f (I x (I x))}\\ &=\text{f (S I I x)}\\ &=\text{(K f x) (S I I x)}\\ &=\text{S (K f) (S I I) x}\\ &=\text{(K S f) (K f) (S I I) x}\\ &=\text{S (K S) K f (S I I) x}\\ &=\text{S (K S) K f (K (S I I) f) x}\\ &=\text{S (S (K S) K)(K (S I I)) f x}\\ \therefore \text{ E}&=\text{S (S (K S) K)(K (S I I)) f}\\ &=\text{T f [Let]} \end{align*} $$

Now $\text{Y}=\lambda\text{f. E E}$, so: $$\begin{align*} \text{Y f}&=\text{E E}\\ &=\text{T f (T f)}\\ &=\text{S T T f}\\ \therefore\text{ Y}&=\text{S T T}\\ &=\text{S S I T}\\ &=\text{S S I (S (S (K S) K)(K (S I I)))}\\ &=\text{S (K (S I I)) (S (S (K S) K)(K (S I I)))} \end{align*}$$

Note: See this for why $\text{SSI}$ and $\text{S(K(SII))}$ are equivalent.

Answer 2

Define $D = S I I$, $B = S (K S) K$ and $V = S B (K D)$. Then, your definition is $Y = S (K D) V$. Since $D x = S I I x = I x (I x) = x x$, then, when applied to $f$, this is equivalent to $$S (K D) V f = K D f (V f) = D (V f) = V f (V f) = S V V f.$$ Thus, under the η-rule, $$S (K D) V = λf·S (K D) Vf = λf·S V V f = S V V.$$ Actually, if you go with strict $SKI$ abstraction, then $S V V$ is what the article should be citing, not $S (K D) V$.

First, $$f(x x) = K f x (D x) = S (K f) D x\quad⇒\quadλx·f(x x) = λx·S (K f) D x = S (K f) D.$$ Second, $$S (K f) = K S f (K f) = S (K S) K f = B f,\quad D = K D f.$$ Therefore $$S (K f) D = B f (K D f) = S B (K D) f = V f.$$ Thus, it follows that $$λf·(λx·f (x x))(λx·f (x x)) = λf·(V f)(V f) = λf·S V V f = S V V.$$

Under the combinator engine Combo, which I put up on GitHub, the abstraction algorithm uses a wider range of combinators, including $D$ and $B$. If will yield $S \_0 \_0$, where $\_0 = C B D$, since $C a b = S a (K b)$ (under the η-rule), where $C = λxλyλz·x z y$. Therefore, $\_0 = V$. If you run Combo on $V = S B (K D)$ with the extensionality axiom (the η-rule) turned on, it will give you $C B D$.

If you run it on $S (C B D) (C B D)$ with, extensionality turned on, it will block it and report it as a "cyclic term", recognizing that it reduces as $Y = λf·Y f = λf·f (Y f)$, which leads to an infinite reduction. If I get around to upgrading Combo to generate, accept and process rational infinite lambda terms, then it might eventually be able to state that $Y$ has $λf·f(f(f(⋯))$ as a normal form, establishing this result in the same finite number of steps that it currently takes to recognize it's a cyclic term in its current version.

Answer 3

$Y$ combinator is defined as: $$\text{Y}\ :=\ \lambda f. (\lambda x.f(xx))\ (\lambda x.f(xx))$$

To convert it to SKI-combinators, first apply lambda lifting: $$ \lambda x.f(xx) = (\lambda f.\lambda x.f(xx))\ f\ =:\ H f$$ Writing in combinators notation, $$ H f x = f (x x) $$ $$ Y f = H f (H f) = f(Hf(Hf)) $$ Now we can convert the $Y$ and $H$ definitions by matching right-hand sides of the combinator equations

$\qquad\begin{align} I a &= a \\ K a b &= a \\ S a b c &= a c (b c) \\ S I I c &= I c (I c) = c c \end{align}$

then,

$\qquad\begin{align} H f x &= f (x x) = K f x (SII x) \\ &= S(Kf)(SII)x \\ &= S(KS)Kf(K(SII)f)x \\&= S(S(KS)K)(K(SII))fx \end{align}$

$\qquad\begin{align} \ \ Y f &= Hf(Hf) = SII(Hf) \\&= K(SII)f(Hf) \\&= S(K(SII))H f \end{align}$

$\qquad\begin{align} \ \ Y f &= Hf(Hf) = SHHf \\&= SSIHf \end{align}$

Substituting the $H$ definition gives us two possibilities, after $\eta$-contraction:

$$ Y = S(K(SII))(S(S(KS)K)(K(SII))) $$ $$ Y = SSI(S(S(KS)K)(K(SII))) $$

Applying the $S$ rule, the first definition can be further contracted to

$$ Y = SS(S(S(KS)K))(K(SII)) $$

which seems to be the shortest encoding of $Y$, also in terms of $SK$-letters only, since each $I$ counts as three (its explicit encoding shortly follows below).

Working with just $SKI$ combinators can fast become unwieldy. Adding more basic combinators helps:

$\quad\begin{align} Babc &= a(bc) = Kac(bc) = S(Ka)bc \\ Cabc &= ac(b) = ac(Kbc) = Sa(Kb)c = S(K(Sa))Kbc \\ Wab &= abb = ab(SKab) = Sa(SKa)b = SS(SK)ab \\ &= abb = ab(Ib) = SaIb = CSIab = SS(KI)ab \\ Ub &= bb = Ibb = WIb = Ib(Ib) = SIIb \\ Ia &= a = Ka(Ka) = SKKa = SKSa \end{align}$

$U$ isn't usually included as a basic combinator, but it is arguably just as fundamental. Now we can have

$\quad\begin{align} H f x &= f (xx) = f (Ux) = BfUx = CBUfx \\ H f x &= f (xx) = Bfxx = W(Bf)x = BWBfx \\ Y f &= H f (H f) = SHHf = SSIHf = WSHf \end{align}$

and thus the previous definitions were actually

$\quad\begin{align} \ \ Y &= S(KU)(SB(KU)) = BU(CBU) \\ \ \ Y &= SSI(SB(KU))) = SSI(CBU) \\ \ \ Y &= SS(SB)(KU) \end{align}$

but now we've found yet other encodings,

$\quad\begin{align} \ \ Y &= WS(CBU) \\ \ \ Y &= BU(BWB) = SSI(BWB) = WS(BWB) \end{align}$

There's also $Y = BU(CBU) = SB(CB)U = SSCBU $, but substituting the definitions in it leads to a very long $SK$-encoding, evidently.

To recap,

$\quad\begin{align} \ \ Y &= S(K(SII))(S(S(KS)K)(K(SII))) \\ \ \ Y &= SSI(S(S(KS)K)(K(SII))) \\ \ \ Y &= SS(S(S(KS)K))(K(SII)) \end{align}$

These are 22(14), 18(12), 15(11) $SK(I)$-letters encodings. They are all different encodings for the same combinator definitions $Yf = Hf(Hf)$ and $Hfx=f(xx)$.

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Using combinators instead of lambda-expressions makes it much easier to come up with a definition for a fixpoint combinator, in the first place: $$ Yf = Hf(Hf) = f(Hf(Hf)) \ \ \models \ \ Hfx = f(xx) $$ $$ Y\text{′} f = HfH = f(HfH) \ \ \models \ \ Hfx = f(xfx) $$ $$ \Theta f = HHf = f(HHf) \ \ \models \ \ Hxf = f(xxf) $$

giving rise to

$$ Y = BU(CBU) = SS(S(S(KS)K))(K(SII)) $$ $$ Y\text{′} = WC(SB(C(WC))) = SSK(S(K(SS(S(SSK))))K) $$ $$ \Theta = U(B(SI)U) = SI(S(K(SI)))(SII) $$

(the $SK$-only encoding for $Y\text{′}$ is due to John Tromp, according to Wikipedia). These are 15(11), 12(12), 17(9) $SK(I)$-letters encodings.

It also holds that $$ \Theta = WI( I( B(SI)(WI)) ) $$ $$ Y\text{′} = WC( C( B(SI)(WC)) ) $$ but this leads to a bit longer, 16(14)-long encoding $$ Y\text{′} = SSK(S(K(S(S(K(SI))(SSK))))K) $$

Answer 4

To demontrate that $\textsf Y=\textsf{S(K(SII))(S(S(KS)K)(K(SII)))}$, introduce those combinators aiming to migrate the lambda bound-variables rightwards until the expression may be $\eta$-reduced; if $\mathsf Y$ is indeed equivalent to that combinator string.

$\qquad\begin{align} \textsf Y &:= \lambda f.(\lambda x.f(xx))(\lambda x.f(xx)) \\ &~= \lambda f.(\lambda x.f(\textsf Ix(\textsf Ix)))(\lambda x.f(\textsf Ix(\textsf Ix)))&&\mathsf I\text{-introduction}\because a=\mathsf Ia \\ &~= \lambda f.(\lambda x.f(\textsf{SII}x))(\lambda x.f(\textsf{SII}x))&&\mathsf S\text{-introduction }\because ac(bc)=\mathsf Sabc \\ &~= \lambda f.(\lambda x.\textsf{K}fx(\textsf{SII}x))(\lambda x.\textsf{K}fx(\textsf{SII}x))&&\mathsf K\text{-introduction }\because a=\mathsf Kab \\ &~= \lambda f.(\lambda x.\textsf{S(K}f\textsf{)(SII)}x)(\lambda x.\textsf{S(K}f\textsf{)(SII)}x)&&\mathsf S\text{-introduction } \\ &~= \lambda f.\textsf{S(K}f\textsf{)(SII)(S(K}f\textsf{)(SII))}&&\eta\text{-reduction} \\ &~= \lambda f.\textsf{KS}f\textsf{(K}f\textsf{)(SII)(KS}f\textsf{(K}f\textsf{)(SII))}&&\mathsf K\text{-introduction } \\ &~= \lambda f.\textsf{S(KS)K}f\textsf{(SII)(S(KS)K}f\textsf{(SII))}&&\mathsf S\text{-introduction } \\ &~= \lambda f.\textsf{S(KS)K}f\textsf{(K(SII)}f\textsf{)(S(KS)K}f\textsf{(K(SII)}f\textsf{))}&&\mathsf K\text{-introduction } \\ &~= \lambda f.\textsf{S(S(KS)K)(K(SII))}f\textsf{(S(S(KS)K)(K(SII))}f\textsf{)}&&\mathsf S\text{-introduction } \\ &~= \lambda f.\textsf{I(S(S(KS)K)(K(SII))}f\textsf{)(I(S(S(KS)K)(K(SII))}f\textsf{))}&&\mathsf I\text{-introduction } \\ &~= \lambda f.\textsf{SII(S(S(KS)K)(K(SII))}f\textsf{)}&&\mathsf S\text{-introduction} \\ &~= \lambda f.\textsf{K(SII)}f\textsf{(S(S(KS)K)(K(SII))}f\textsf{)}&&\mathsf K\text{-introduction } \\ &~= \lambda f.\textsf{S(K(SII))(S(S(KS)K)(K(SII)))}f&&\mathsf S\text{-introduction} \\ &~= \textsf{S(K(SII))(S(S(KS)K)(K(SII)))}&&\eta\text{-reduction} \end{align}$

Alternatively, run through those steps in reverse order to demonstrate that the combinator string $\beta$-reduces to the definition for $\sf Y$.