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S combinator always duplicates its last parameter, never deletes it. That's why K is needed for universality.

This can be proved by induction. Or you can cite Craig's theorem (the less known one) for that. See [1]

Honestly, I don't see the endgame here.

[1] https://math.stackexchange.com/questions/839926/is-there-a-p...

That doesn't matter. You could imagine a system that accumulates two terms: (actual result, junk). Instead of deleting something it just adds it to the junk part of the pair. Maybe the junk part itself has computation which never ends, but it doesn't matter because you just extract your result from the left part of the pair.

cvoss, ezwoodland, tromp and v64 made a good point. As v64 points I was thinking of a Combinatory Completeness not Turing Completeness. Dropping that requirement as cvoss, ezwoodland, tromp point you can simulate deletion (that's what quantum computing does btw see No-deleting theorem).

I see the endgame now, thanks guys.

I don't follow your argument. Why must we have the ability to "delete" sub-expressions?

Consider a computational model that, rather than work by successively rewriting an expression over and over in a way that honors some equivalence relation over expressions, it works by explicitly building the sequence of such expressions. In that kind of system, every computational state properly contains the previous state. Things grow and grow and never get "deleted". Yet such a system can clearly be universal.

as far as I understand it, turing completeness is a weaker property than combinatory completeness, which Craig's theorem is addressing. The nonexistence of a singleton combinatory basis doesn't necessarily imply the nonexistence of a turing complete combinator.

An implicit K suffices for universality, as in \x\y\z. x z (y (\w.z))

Barendregt & Manzonetto's 2022 "A lambda calculus satellite" has a whole chapter on the S fragment, for those interested

Wait wouldn't this revolutionize computing? Seems like a rather low bounty for such a monumental proof

No? Why would it?

In the negative case, it would say the idea doesn't pan out.

In the positive case, it would mean that you can use just S instead of S and K when doing combinator reduction, but doesn't change that this kind of reduction is not super efficient practically speaking.

I think that website cost more than the listed prize amount.

Coincidental timing for Wolfram to pop up here just as it becomes clearer that he might actually have met Epstein after all.

More wolf-slop.