NO

Problem:
 a(x) -> b(b(b(x)))
 a(x) -> c(x)
 b(x) -> b(x)

Proof:
 Nonconfluence Processor:
  terms: c(f61()) *<- a(f61()) ->* b(b(b(f61())))
  Qed
  
  first automaton:
   final states: {1}
   transitions:
    c(2) -> 1*
    f61() -> 2*
  
  second automaton:
   final states: {3}
   transitions:
    b(4) -> 5*
    b(5) -> 6*
    b(6) -> 3*
    f61() -> 4*