NO

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

Proof:
 Nonconfluence Processor:
  terms: f(h(a())) *<- c() ->* f(f(h(a())))
  Qed
  
  first automaton:
   final states: {1}
   transitions:
    f(3) -> 1*
    a() -> 2*
    h(2) -> 3*
    h(1) -> 1*
  
  second automaton:
   final states: {9}
   transitions:
    f(12) -> 9*
    f(11) -> 12*
    a() -> 10*
    h(10) -> 11*
    h(12) -> 12*
    h(9) -> 9*