Okay. We are proving the optimality of the Grover's algorithm, and to do this, we need to find some vector phi for which this double inequality is true. We already proved that for the right part of an inequality holds for any unitary vector phi, and it's time to prove the left part. Here remember how decomposition of some very effective to most effective algorithm provided by some algorithm designer, and the operators U Omega here requires an oracle, and these separators U1, U2, etc upto Ut are the operators provided by the algorithm designer. Now, we are going to reduce this vector phi t, which is defined like this. We get this most effective algorithm. Instead of all the oracle queries, we put the identity operators, and this trimmed algorithm, we apply to the initial state of the system vector psi zero. After t iterations, we get this vector phi t. Now we are going to introduce some auxiliary values. The first one is this E of omega and T. The E stands for error, because our trimmed algorithm is obviously erroneous, it doesn't call the oracle. Instead of an oracle, it calls the identity operators, so we are going to estimate how much deviation from the correct path we get on each step. So let us recall what is this Q omega operator. It is identity minus two omega, omega, and minus this identity, and everything is applied to psi t. Now, this identities go away, and what we get, we get just two scalar products of omega and psi t. Now we are going to define this function f of t, which is the deviation of the correct vector and the vector provided by the trimmed algorithm on the step t. Let's rewrite it for the previous step. Ut U omega to psi t minus one, minus Ut phi t minus one. Since the operator Ut is unitary, I can remove it from under the norm, and I'm going to add and subtract this value under the null. Let's see what it gives us. So we have U omega psi t minus one, minus psi t minus one, plus psi t minus one, minus phi t, minus one. This is our error here, and this is f of t minus one. So we use the triangle inequality to get this. Okay and for the step t, we're going to use this recursive definition that we get and this estimation for an error. So we have this two from here, and the sum from t equal to one up to T, is modulus omega psi t minus one. Okay. We get this, and for the inequality, we have to put squares here. Because in the middle part of inequality, there is a square. So we get this, and finally, it looks like this. Now, you remember we have proved this lemma in the previous episode, in the previous lecture. Since we proved it once, we can use it as many times as we want. So we do, we use it, and this appears to be less or equal, and 4T, and the sum here of this omega psi t minus one squared, and we don't need a brace here. Just let me think if I put everything correct. I think I did, and you remember, in the middle part of inequality, we have the sum over all vectors of the space. So we sum it over all Omegas, and they get this expression here, this one we just obtained in the previous slide. I'm going to change the places of these sums here. So this is equal to 4T, sum over t, then sum over all omega. All this dot product squared. You can see that this sum here is actually the norm of this vector psi t minus one, which is actually one. So we have t runs here from one to T beak. We have T beak ones and it 4T squared. We just proved the left side of our double inequality for the vector phi t.