NO

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

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