NO

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

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