NO

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

Proof:
 Nonconfluence Processor:
  terms: a(b(c(x37))) *<- a(a(a(x37))) ->* b(c(a(x37)))
  Qed