MAYBE

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

Proof:
 Open