NO

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

Proof:
 Nonconfluence Processor:
  terms: b(a(b(a(b(x10))))) *<- b(a(a(b(a(a(x10)))))) ->* a(a(x10))
  Qed