NO

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

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