MAYBE

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

Proof:
 Open