MAYBE

Problem:
 c(b(x)) -> a(c(x))
 b(a(x)) -> c(c(x))
 b(c(x)) -> c(a(x))
 c(c(x)) -> a(b(x))
 b(c(x)) -> a(c(x))
 c(a(x)) -> b(c(x))
 b(b(x)) -> c(a(x))
 a(a(x)) -> a(b(x))

Proof:
 Open