NO

Problem:
 b() -> f(a())
 a() -> c()
 a() -> a()
 h(c(),f(f(f(a())))) -> h(f(f(c())),h(b(),a()))
 f(f(f(f(h(h(c(),h(f(c()),f(f(h(h(h(h(f(a()),a()),c()),a()),h(h(h(b(),a()),
                                                                f(h(f(h(f(b()),f(f(b())))),a()))),
                                                              h(c(),f
                                                                    (
                                                                    h
                                                                    (
                                                                    b(),f(f(h(h(c(),f(a())),a())))))))))))),
           f(f(b())))))))
 -> c()

Proof:
 Nonconfluence Processor:
  terms: h(c(),f(f(f(c())))) *<- h(c(),f(f(f(a())))) ->* h(f(f(c())),h(b(),a()))
  Qed