NO

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

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