NO

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

Proof:
 Nonconfluence Processor:
  terms: f(h(f(f(b())),a())) *<- f(h(f(b()),a())) ->* b()
  Qed
  
  first automaton:
   final states: {1}
   transitions:
    h(5,2) -> 6*
    b() -> 3*
    f(4) -> 5*
    f(6) -> 1*
    f(3) -> 3,4
    a() -> 2*
  
  second automaton:
   final states: {7}
   transitions:
    b() -> 7*
    f(7) -> 7*