NO

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

Proof:
 Nonconfluence Processor:
  terms: b(b(f3())) *<- b(b(c(c(f3())))) ->* c(a(c(a(a(c(f3()))))))
  Qed
  
  first automaton:
   final states: {3}
   transitions:
    f3() -> 4*
    b(4) -> 5*
    b(5) -> 3*
  
  second automaton:
   final states: {6}
   transitions:
    f3() -> 7*
    a(47) -> 148*
    a(7) -> 46*
    a(8) -> 9*
    a(150) -> 151*
    a(46) -> 47*
    a(9) -> 10*
    a(11) -> 12*
    a(148) -> 149*
    a(48) -> 49*
    b(46) -> 47*
    b(8) -> 9*
    b(9) -> 10*
    b(7) -> 46*
    b(11) -> 12*
    b(150) -> 151*
    b(148) -> 149*
    b(47) -> 148*
    b(48) -> 49*
    c(7) -> 8*
    c(12) -> 6*
    c(151) -> 12*
    c(49) -> 10*
    c(149) -> 150*
    c(47) -> 48*
    c(10) -> 11*
    151 -> 6*
    49 -> 11*