NO

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

Proof:
 Nonconfluence Processor:
  terms: c(a(b(f15()))) *<- c(f15()) ->* c(a(f15()))
  Qed
  
  first automaton:
   final states: {15}
   transitions:
    b(16) -> 17*
    b(18) -> 18*
    a(18) -> 18*
    a(17) -> 18*
    f15() -> 16*
    c(18) -> 15*
  
  second automaton:
   final states: {4}
   transitions:
    b(6) -> 6*
    a(5) -> 6*
    a(6) -> 6*
    f15() -> 5*
    c(6) -> 4*