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Set up a recursive substitution to recognise functional equivalence of combinators (so that it could identify ``SKK and ``SKS as functionally equivalent).
Equivalent to the halting problem in the general case. Let an SK-expression simulate a Turing machine M on input x. Is it equivalent to the SK-expression that always returns true?

Ah, now I see what I had wrong.

Stenlund mentions in “Combinators, λ-terms, and proof theory” (and I get the impression from the historical section of Introduction to higher-order categorical logic that this dates back to Curry) a finite equational axiomatisation of what I'm calling functional equivalence (and he calls extensional, or η-, equality), namely:

```Sxyz = ``xz`yz
``Kxy = x
`Ix = x
`S`KI = I
``BS`S`KK = K
```B`B``B`BSSBS = ``P```B`BSBSS
``B``S``BBS`KKK = `BK
`S`KI = ``SB`KI
where he takes I as a primitive combinator, B = ``S`KSK is as above, and ````Pfgxy = ``f`gx`gy. (One usually writes Ψ instead of P, but I don't know how to get entities inside code tags. :-) )

However: Having a finite axiomatisation of a theory is not the same as being able to decide the truth of an arbitrary proposition in that theory—witness Peano arithmetic—and that, of course, is where my ambition smacks against your refusal to solve the halting problem. :-)

In reply to Re^2: Turing completeness and regular expressions by JadeNB
in thread Turing completeness and regular expressions by JadeNB

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