MAYBE

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

Proof:
 Open