NO

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

Proof:
 Nonconfluence Processor:
  terms: b() *<- f(a()) ->* c()
  Qed
  
  first automaton:
   final states: {8}
   transitions:
    b() -> 8*
    f(8) -> 22*
    f(22) -> 8*
    h(48,8) -> 22*
    a() -> 48*
  
  second automaton:
   final states: {9}
   transitions:
    c() -> 9*