NO

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

Proof:
 Nonconfluence Processor:
  terms: a(b(b(c(x13)))) *<- a(a(b(a(x13)))) ->* a(b(c(a(x13))))
  Qed