MAYBE

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

Proof:
 Open