SK calculus is Turing-complete.[1]
A computational system is Turing-complete if it can simulate an universal Turing machine.
We now simulate a call-by-value SK calculus in Scheme, a dialect of Lisp: (define (((S x) y) z) ((x z) (y z)))
(define ((K x) y) x)
Therefore, Lisp is Turing-complete.
[1] Any λ-calculus expression can be converted into S and K by using the following equivalences: (λx.M) = (KM), if x does not occur free in M. (λx.x) = (SKK) (λx.MN) = (S(λx.M)(λx.N))
λ-calculus is Turing-complete. Proving it is left as an exercise for the reader.