NO

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

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