NO

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

Proof:
 Nonconfluence Processor:
  terms: h(f(h(a()))) *<- c() ->* f(f(h(a())))
  Qed
  
  first automaton:
   final states: {1}
   transitions:
    h(4) -> 4,1
    h(2) -> 3*
    f(3) -> 4*
    a() -> 2*
  
  second automaton:
   final states: {8}
   transitions:
    h(11) -> 11*
    h(8) -> 8*
    h(9) -> 10*
    f(11) -> 8*
    f(10) -> 11*
    a() -> 9*