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Specifically, if I defined a new $K_2$ as $$K_2 = \lambda x. (\lambda y. y)$$ instead of $$K = \lambda x. (\lambda y. x)$$ would the $\{S, K_2,I\}$-calculus be a compete basis?

My guess is "no," just because I can't seem to be able to construct the regular K combinator from the $S$, $I$, and $K_2$ combinators, but I don't have an algorithm to follow, nor do I have good intuition about making things out of these combinators.

It seems like you can define $$K_2 = K I$$ with the regular $\{S, K, (I)\}$-calculus, but I couldn't really work backwards from that to get a derivation of $K$ in terms of $K_2$ and the rest.

My attempt at a proof that it was not functionally complete essentially attempted to exhaustively construct every function attainable from these combinators in order to show that you reach a dead end (a function you've seen before) no matter what combinators you use. I realize that this isn't necessarily going to be true of functionally incomplete sets of combinators (e.g. the $K$ combinator on its own will never dead end when applied to itself), but this was my best thought. I was always able to use the $S$ combinator to sneak out of what I thought was finally a dead end, so I'm no longer so sure of the feasibility of this approach.

I asked this question on StackOverflow but was encouraged to post it here. I received a few comments on that post, but I'm not sure I understood them right.

Bonus: if it isn't a complete basis, is the resulting language nonetheless Turing-complete?

Anton Trunov
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cole
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2 Answers2

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Consider the terms of the $S,K_2,I$ calculus as trees (with binary nodes representing applications, and $S, K_2$ leaves representing the combinators.

For example, the term $S(SS)K_2$ would be represented by the tree

        @
       / \
      /   \
     @    K2
    / \
   /   \
  S     @
       / \
      /   \
     S     S

To each tree $T$ associate its rightmost leaf, the one you get by taking the right branch at each @. For example, the rightmost leaf of the tree above is $K_2$.

As can be seen from the ASCII art below, all reduction rules in the $S, K_2, I$ calculus preserve the rightmost leaf.

         @                           @
        / \                         / \
       /   \                       /   \
      @     g    [reduces to]     @     @
     / \                         / \   / \
    /   \                       e   g f   g
   @     f                 
  / \
 /   \
S     e

      @
     / \
    /   \
   @     f    [reduces to]   f
  / \
 /   \
K2    e

From there on, it's easy to see that if some term $T$ reduces to $T'$, then $T$ and $T'$ have the same rightmost leaf. Hence, there is no term $T$ in the $S, K_2, I$ calculus such that $TK_2S$ reduces to $K_2$. However, $KK_2S$ reduces to $K_2$, hence $K$ cannot be expressed in the $S,K_2, I$ calculus.

Z. A. K.
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EDIT: As the comments point out, this is only a partial answer, since it applies only to the simply-typed $S,K_2,I$ calculus (or rather, it shows that there is no possible definition of K that does not contain an ill-typed subterm). If there's no objection, I won't delete it, since it presents a very productive proof technique for the typed setting.

Recall that our combinators have the following types (Curry-style, so $A,B,C$ are variable):

  • $K: A \rightarrow B \rightarrow A$
  • $K_2: A \rightarrow B \rightarrow B$
  • $S: (A \rightarrow B \rightarrow C) \rightarrow (A \rightarrow B) \rightarrow (A \rightarrow C)$
  • $I: A \rightarrow A$

By the Curry-Howard correspondence, if we can express $K$ in terms of $I,S,K_2$ then the Hilbert-style proof calculus with three logical axioms $A \rightarrow B \rightarrow B, (A \rightarrow B \rightarrow C) \rightarrow (A \rightarrow B) \rightarrow (A \rightarrow C), A \rightarrow A$ and one inference rule (from $A$ and $A \rightarrow B$ infer $B$) proves the formula $A \rightarrow B \rightarrow A$.

But we can give a three-valued (values t,f,u) semantics to the connective $\rightarrow$ such that the formulas $A \rightarrow B \rightarrow B$, $(A \rightarrow B \rightarrow C) \rightarrow (A \rightarrow B) \rightarrow (A \rightarrow C)$ and $A \rightarrow A$ get the value t for any interpretation.

A B | A -> B
t t | t
t f | f
f t | t
f f | t
t u | f
f u | t
u t | t
u f | f
u u | t

This semantics is clearly sound in the sense that every consequence of the axioms $K_2, S, I$ gets the value t under every interpretation (it is not complete, there are things that always get the value t but that we cannot actually prove). However, $A \rightarrow B \rightarrow A$ gets the value f under the interpretation that assigns u to $A$ and t to $B$, and is therefore not provable from the axioms corresponding to $S, K_2, I$.

Z. A. K.
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