NO

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

Proof:
 Nonconfluence Processor:
  terms: h(f(c())) *<- a() ->* h(g(c()))
  Qed
  
  first automaton:
   final states: {8}
   transitions:
    h(10) -> 8*
    h(53) -> 10*
    c() -> 9*
    f(9) -> 10*
    g(10) -> 10*
    g(9) -> 53*
    g(53) -> 53*
  
  second automaton:
   final states: {14}
   transitions:
    h(16) -> 14*
    c() -> 15*
    g(15) -> 16*
    g(16) -> 16*