NO

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

Proof:
 Nonconfluence Processor:
  terms: b(b(x109)) *<- b(b(c(c(x109)))) ->* b(c(b(b(b(b(x109))))))
  Qed