NO

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

Proof:
 Unfolding Processor:
  loop length: 6
  terms:
   a(b(b(b(b(b(b(x887)))))))
   b(b(a(a(b(b(b(b(x887))))))))
   b(b(a(b(b(a(a(b(b(x887)))))))))
   b(b(a(b(b(a(b(b(a(a(x887))))))))))
   b(b(a(b(b(b(b(a(a(a(a(x887)))))))))))
   b(b(a(b(b(b(b(b(a(a(x887))))))))))
  context: b(b([]))
  substitution:
   x887 -> x887
  Qed