NO

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

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