NO

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

Proof:
 Nonconfluence Processor:
  terms: f(c()) *<- f(a()) ->* b()
  Qed
  
  first automaton:
   final states: {7}
   transitions:
    c() -> 8*
    f(8) -> 7*
  
  second automaton:
   final states: {6}
   transitions:
    b() -> 6*
    a() -> 14*
    c() -> 14,15
    h(6,16) -> 6*
    h(15,14) -> 16*