NO

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

Proof:
 Nonconfluence Processor:
  terms: b(b(c(x29))) *<- b(a(a(x29))) ->* a(x29)
  Qed