NO

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

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