NO

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

Proof:
 Nonconfluence Processor:
  terms: f(g(g(f(g(f3()))))) *<- f(g(f(g(f(f3()))))) ->* g(f(g(g(f(f3())))))
  Qed
  
  first automaton:
   final states: {8}
   transitions:
    f3() -> 9*
    f(13) -> 8*
    f(10) -> 11*
    g(12) -> 13*
    g(9) -> 10*
    g(11) -> 12*
  
  second automaton:
   final states: {14}
   transitions:
    f3() -> 15*
    f(15) -> 16*
    f(18) -> 19*
    g(19) -> 14*
    g(16) -> 17*
    g(17) -> 18*