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May 2, 2019 at 23:28 history edited Z. A. K. CC BY-SA 4.0
Edited according to the comments to show that this is only a partial answer
May 2, 2019 at 22:49 comment added Z. A. K. Charles Stewart's observation is correct. While it is true that every well-typed proper combinator that can be defined in the SKI calculus can also be defined in the simply-typed SKI calculus, the same thing need not be (a priori) true of the SK2I calculus.
May 2, 2019 at 19:33 comment added cole I’m afraid that I’m a little in over my head here, so I’m going to remove my acceptance of this answer and update my question to indicate that I’d be fine with either an argument that it’s Turing (in)complete or an (in)complete basis. I’ll try to check in later to see if I should leave the question unaccepted.
May 2, 2019 at 17:44 comment added chi @NoamZeilberger I also agree. This argument unfortunately seems insufficient: in the untyped calculus you might be able to obtain $K$ using a construction which involves (sub)terms which have no type. One should prove that if you can achieve that, you can also do the same using only typed terms, but that looks hard to prove to me.
May 2, 2019 at 16:08 comment added Noam Zeilberger I agree with Charles Stewart (over here: twitter.com/txtpf/status/1123962607306706944) that it's not clear how to pass from uninhabitation in simply-typed lambda calculus to inexpressibility using combinators. There might be an argument specific for K, but the initial step "...then one could also do the same thing in the simply-typed λ-calculus" doesn't hold in general (Charles mentioned the counterexample of the Y combinator). Do you see of making this argument rigorous?
May 1, 2019 at 21:53 vote accept cole
May 2, 2019 at 19:33
May 1, 2019 at 18:13 history edited Z. A. K. CC BY-SA 4.0
clarified language
May 1, 2019 at 16:34 history edited Z. A. K. CC BY-SA 4.0
minor clarifications
May 1, 2019 at 16:31 comment added Z. A. K. @robin-houston: please see my edit (I've also added a different, semantic argument with the same conclusion).
May 1, 2019 at 16:02 history edited Z. A. K. CC BY-SA 4.0
Added a simple semantic proof, commented on missing parts of original argument.
May 1, 2019 at 13:07 comment added Robin Houston Can you sketch how to prove S in this restricted sequent calculus? It doesn’t seem to be possible with the rules I guessed you might mean.
May 1, 2019 at 12:00 comment added Noam Zeilberger I like the approach, but could you clarify what rules you are taking as your sequent calculus?
May 1, 2019 at 11:07 history edited Z. A. K. CC BY-SA 4.0
corrected typos, slight rewording, formatting
May 1, 2019 at 11:05 review First posts
May 1, 2019 at 12:38
May 1, 2019 at 11:02 history answered Z. A. K. CC BY-SA 4.0