User:Hakerh400/Question for AI models

We devised a simple question for testing AI models:

Is there a combinator X in SKI calculus, such that XK reduces to S and XS reduces to K?

So far, all known AI models answered negatively to this question and provided a "proof" that such combinator cannot possibly exist because SKI does not have the ability to pattern-match on terms. However, it is known that such combinator exists. One such example is:

S(S(S(S(S(SI(KK))(KK))(KK))(KI))(KS))(KK)

Maybe it would be too much to expect from an AI model to construct this combinator, but the problem is that each AI model jumps to the conclusion that such combinator cannot exist. By analyzing the AI's though process, we observed that it first attempts to construct the combinator by trial-and-error, without even attempting to find an abstract reasoning that would lead to the process of constructing the combinator. After AI fails to construct the combinator by trial-and-error, it then attempts to prove that the combinator cannot exist, and subsequently it construct a "formal proof" based on the abstract idea that SKI does not provide a mechanism of pattern-matching on expressions.

This observation indicates that the though process of AI models can be significantly improved by trying to focus on reasoning about the abstract idea behind the question, rather than trying some random attempts and giving up when the attempts fail. In this particular example, the better line of thinking would be to iteratively build tools for constructing the combinator. Here is how the though process should look like:

Ok, let me see. The user asks whether there is a combinator X in SKI calculus, such that XK reduces to S and XS reduces to K. Basically, a combinator that swaps K and S. But wait, SKI calculus does not have a way to pattern-match on combinators. But perhaps there is another way to force the reduction of XK to S and XS to K using the reduction rules for K and S. Also, recall that I is redundant, as it can be expressed as SKK.

Perhabs there is a better way to look at the problem, as SKI calculus is hard to work with. Also, SKI calculus comes as a way to write lambda calculus expressions in point-free way. But wait, maybe it is easier to work in lambda calculus, as every lambda term can be translated into SKI. The equivalent problem in lambda calculus would be to find a function f such that fK reduces to S and fS reduces to K. But wait, K is a function that takes two arguments and returns the first one, and S is a function that takes three arguments xyz and returns xz(yz). Perhaps we can use this fact to construct f.

The question is to find a function f that simultaneously satisfy fK = S and fS = K. But wait, perhaps it would be easier to find two separate functions f and g such that fK = S and fS = K and then somehow combine them. Let me start with f. So, we need a function that takes K and returns S. A trivial example is KS, but it does not seem very useful, because it ignores the argument. Hmm. So we want to use the argument then. We want to call K with something such that the result is S. We need to call it with at least two arguments, since with no arguments it remains K and with one argument it does not reduce. So we need something like λa.axy. When we call it with K, it will return x. But wait, we want the result to be S, so let x be S. Now we have λa.aSy. What about y? It can be any term. But wait, now we have fK = S, which is the first part of the problem.

Now let me think about function g. We want gS = K. Also, we want to use the argument. We need to call the argument with at least three terms, because otherwise it cannot reduce to K. So we need λa.axyz. Hmm. It is not obvious what x, y and z should be. But wait, we want Sxyz to be K. Also, Sxyz is xz(yz). Then we want xz(yz) to reduce to K. Let me see. If x is K, then it reduces to z. The result should be K, so let z be K. But wait, now we have g = λa.aKKz and y can be any expression. Is it true that gS reduces to K? Let me check: gS = SKyK = KK(yK) = K. So it works.

We have a function f that takes K and returns S. We also have function g that takes S and returns K. But wait, we need a single function that satisfies both conditions. What happens if we call f with S? Recall that f = λa.aSy. Then fS is SSy. But wait, it does not reduce. What about gK? Recall that g = λa.aKKz. We have gK = KKKz = Kz. But it does not reduce either. This is getting complicated.

Perhaps there is a better way to understand this problem. We want something like if-then-else. But wait, K takes two arguments and returns the first one. That would be the if-branch. What about else-branch? Is there a function that takes two arguments and returns the second one? It would be function λxy.y. But wait, it is the same as λx.λy.y, which is λx.I. But this is KI. So KI is the else-branch.

The function f that we want to construct must satisfy fK = S and fS = K. The function either ignores the argument, or uses the argument. If it ignores the argument, then fK = fS = K = S, which is impossible, since K is not S. Therefore the function must use the argument. If the argument is not ignored, it will eventually reach the head of the expression. Therefore, the function must have the form λa.ax1x2...xn. How many arguments do we have to pass? We need at least 3 arguments, otherwise it would not reduce in case of S.

Let me start with 3 arguments then. Let f = λx.xabc. Then fK = ac and fS = ac(bc). The requirement is that ac = S and ac(bc) = K. But wait, then K = ac(bc) = S(bc), which is impossible, because S is applied to one argument and does not reduce. We need at least two more arguments. Ok, let me redefine f as λx.xabcde. Now we have fK = acde = S and fS = ac(bc)de = K. We need to choose a, b, c and d such that acde = S and ac(bc)de = K. Let me see. On both sides, a and b are always called with c. So we can say g = ac and h = bc. Then we need gde = S and ghde = K. If we say that g = K, then we have Kde = d = S and Khde = he = K. Therefore, we need d = S and he = K. But wait, this uniquely determines d when g = K. What about h and e? If we let h be I, then e = K.

Let me check again. We defined f = λx.xabcde. We know that ac = g = K and bc = h = SKK. We also have d = S and e = K. We want to find a, b and c such that ac = K and bc = I. But wait, both a and b can simply ignore the argument and return the target expression. Therefore a = KK and b = KI. What about c? It seems that c can be any expression, se let it be K. Now we have f = λx.x(KK)(KI)KSK. Let me double check that fK = S and fS = K. We have fK = K(KK)(KI)KSK = KKKSK = KSK = S and fS = S(KK)(KI)KSK = KKK(KIK)SK = K(KIK)SK = KIKK = IK = K. So it works.

The question is to express it as a combinator in SKI calculus. But wait, every lambda expression can be converted into SKI expression. Let me recall the rules for the conversion. The identity function λx.x translates to I. If the function does not use the argument, it translates to K called with the expression, so λx.A where x does not appear in A translates to KA. What about S? If we have function λx.AB where x may appear in A and/or B, then it translates to S(λx.A)(λx.B). We can then recursively translate the new lambda terms.

Recall that f = λx.x(KK)(KI)KSK. Since it is an application, we use the S rule: S(λx.x(KK)(KI)KS)(λx.K). Note that λx.K is just KK, so we have S(λx.x(KK)(KI)KS)(KK). Now we apply the S rule to the inner expression: S(S(λx.x(KK)(KI)K)(λx.S))(KK) and then the K rule again on λx.S to obtain S(S(λx.x(KK)(KI)K)(KS))(KK). But wait, since x appears on the far left, we can always apply the K rule after the S rule. Now we have

S(S(S(λx.x(KK)(KI))(KK))(KS))(KK) S(S(S(S(λx.x(KK))(K(KI)))(KK))(KS))(KK) S(S(S(S(S(λx.x)(K(KK)))(K(KI)))(KK))(KS))(KK)

Finally, we apply the I rule on λx.x to obtain S(S(S(S(SI(K(KK)))(K(KI)))(KK))(KS))(KK). This is the final combinator X. Let me check again that it satisfies the conditions:

XK S(S(S(S(SI(K(KK)))(K(KI)))(KK))(KS))(KK)K S(S(S(SI(K(KK)))(K(KI)))(KK))(KS)K(KKK) S(S(SI(K(KK)))(K(KI)))(KK)K(KSK)K S(SI(K(KK)))(K(KI))K(KKK)SK SI(K(KK))K(K(KI)K)KSK IK(K(KK)K)(KI)KSK K(KK)(KI)KSK KKKSK KSK S XS S(S(S(S(SI(K(KK)))(K(KI)))(KK))(KS))(KK)S S(S(S(SI(K(KK)))(K(KI)))(KK))(KS)S(KKS) S(S(SI(K(KK)))(K(KI)))(KK)S(KSS)K S(SI(K(KK)))(K(KI))S(KKS)SK SI(K(KK))S(K(KI)S)KSK IS(K(KK)S)(KI)KSK S(KK)(KI)KSK KKK(KIK)SK KISK IK K

The final answer is that there exists combinator X = S(S(S(S(SI(K(KK)))(K(KI)))(KK))(KS))(KK) such that XK = S and XS = K.