NO

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

Proof:
 Nonconfluence Processor:
  terms: b(a(b(a(f7())))) *<- b(b(b(f7()))) ->* a(b(a(b(f7()))))
  Qed
  
  first automaton:
   final states: {1}
   transitions:
    a(4) -> 5*
    a(2) -> 3*
    b(5) -> 1*
    b(3) -> 4*
    f7() -> 2*
  
  second automaton:
   final states: {6}
   transitions:
    a(10) -> 6*
    a(8) -> 9*
    b(9) -> 10*
    b(7) -> 8*
    f7() -> 7*