To prove SKK and II are beta equivalent, lambda calculus
I am new to lambda开发者_如何学C calculus and struggling to prove the following.
SKK and II are beta equivalent.
where
S = lambda xyz.xz(yz) K = lambda xy.x I = lambda x.x
I tried to beta reduce SKK by opening it up, but got nowhere, it becomes to messy. Dont think SKK can be reduced further without expanding S, K.
SKK
= (λxyz.xz(yz))KK
→ λz.Kz(Kz) (in two steps actually, for the two parameters)
Kz
= (λxy.x)z
→ λy.z
λz.Kz(Kz)
→ λz.(λy.z)(λy.z) (again, several steps)
→ λz.z
= I
(You should be able to prove that II → I
)
;another approach with fewer steps, first reduce SK to λyz.z;
SKK
= (λxyz.xz(yz))KK
→ λyz.Kz(yz) K
→ λyz.(λxy.x)z(yz) K
→ λyz.(λy.z)(yz) K
→ λyz.z K
→ λz.z
= I
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