It is, though that's not very surprising. Here's a quick proof: S-K combinators[1] are a Turing-complete subset of lambda calculus. Here's an implementation of

It is, though that's not very surprising. Here's a quick proof: S-K combinators[1] are a Turing-complete subset of lambda calculus. Here's an implementation of them: k = (x) -> (y) -> x s = (x) -> (y) -> (z) -> x(z)(y(z)) 1: http://en.wikipedia.org/wiki/SKI_combinator_calculus http://en.wikipedia.org/wiki/SKI_combinator_calculus