NO

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

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