NO

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

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