NO

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

Proof:
 Nonconfluence Processor:
  terms: a(b(a(b(x15)))) *<- a(a(a(a(x15)))) ->* b(a(b(a(x15))))
  Qed