User blog comment:Edwin Shade/A Question About the SKI-Calculus/@comment-30754445-20171108013616

If you're just using combinators that shuffle and regroup things, the answer is yes: None of these systems can be any more powerful than SKI, which is equivalent in strength to the Busy Beaver function.

The reason simple:

SKI is Turing Complete. This means that you can model any computable process with a SKI string.

And all of systems you are talking about, which just shuffle and regroup items, can be easily modeled by a computer program... which, in turn, can be modeled with a SKI string.

To go beyond this, you need a combinator that does something which cannot be solved algorithmically. That is, you need a combinator that does something beyond shuffling, regrouping and deleting items.

One example for this is the Ω combinator from the Xi function. And this is why the Xi function is stronger than ordinary SKI calculus.