NO

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

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