MAYBE

Problem:
 a(a(x1)) -> b(a(b(x1)))
 b(b(x1)) -> a(c(b(x1)))
 c(c(x1)) -> c(b(a(x1)))
 a(b(x1)) -> b(a(x1))
 b(c(x1)) -> c(x1)

Proof:
 Open