NO

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

Proof:
 Nonconfluence Processor:
  terms: a(b(b(b(f2())))) *<- a(a(a(f2()))) ->* b(b(b(a(f2()))))
  Qed
  
  first automaton:
   final states: {37}
   transitions:
    a(41) -> 37*
    b(39) -> 40*
    b(40) -> 41*
    b(38) -> 39*
    f2() -> 38*
  
  second automaton:
   final states: {42}
   transitions:
    a(43) -> 44*
    b(44) -> 45*
    b(46) -> 42*
    b(45) -> 46*
    f2() -> 43*