NO

Problem:
 W(B(x)) -> I(x)
 B(S(x)) -> S(x)
 W(x) -> I(x)

Proof:
 Nonconfluence Processor:
  terms: I(B(f6())) *<- W(B(f6())) ->* I(f6())
  Qed
  
  first automaton:
   final states: {1}
   transitions:
    f6() -> 2*
    B(2) -> 3*
    I(3) -> 1*
  
  second automaton:
   final states: {4}
   transitions:
    f6() -> 5*
    I(5) -> 4*