NO

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

Proof:
 Unfolding Processor:
  loop length: 6
  terms:
   a(a(a(b(a(a(x651))))))
   b(a(b(b(a(a(x651))))))
   a(b(a(b(a(a(x651))))))
   a(a(b(a(a(a(x651))))))
   a(a(b(b(a(b(x651))))))
   a(a(b(a(b(a(x651))))))
  context: []
  substitution:
   x651 -> x651
  Qed