NO

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

Proof:
 Unfolding Processor:
  loop length: 6
  terms:
   a(a(b(b(a(b(b(a(x664))))))))
   b(a(b(b(b(a(b(b(a(x664)))))))))
   b(a(a(a(a(b(b(a(x664))))))))
   b(a(a(b(a(b(b(b(a(x664)))))))))
   b(a(a(b(a(a(a(a(x664))))))))
   b(a(a(b(b(a(b(a(a(x664)))))))))
  context: b([])
  substitution:
   x664 -> b(x664)
  Qed