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(x1043)))))))
   b(b(a(a(b(b(b(b(x1043))))))))
   b(b(a(b(b(a(a(b(b(x1043)))))))))
   b(b(a(b(b(a(b(b(a(a(x1043))))))))))
   b(b(a(b(b(b(b(a(a(a(a(x1043)))))))))))
   b(b(a(b(b(b(b(b(a(a(x1043))))))))))
  context: b(b([]))
  substitution:
   x1043 -> x1043
  Qed