NO

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

Proof:
 Nonconfluence Processor:
  terms: c(e(x140)) *<- d(d(x140)) ->* x140
  Qed