NO

Problem:
 b(a(a(x1))) -> a(x1)
 a(a(a(x1))) -> b(b(b(x1)))
 b(b(x1)) -> a(b(a(x1)))

Proof:
 Unfolding Processor:
  loop length: 6
  terms:
   a(a(a(b(a(a(x2855))))))
   b(b(b(b(a(a(x2855))))))
   a(b(a(b(b(a(a(x2855)))))))
   a(b(a(a(b(a(a(a(x2855))))))))
   a(b(a(a(b(b(b(b(x2855))))))))
   a(b(a(a(a(b(a(b(b(x2855)))))))))
  context: a(b([]))
  substitution:
   x2855 -> b(a(x2855))
  Qed