MAYBE

Problem:
 a(x1) -> b(b(x1))
 c(b(x1)) -> d(x1)
 e(b(x1)) -> c(c(x1))
 d(b(x1)) -> b(f(x1))
 f(x1) -> a(e(x1))
 c(x1) -> x1
 a(a(x1)) -> f(x1)

Proof:
 Open