Hello, if I understand correctly, you are thinking |y> as a single bit state in eq 10. It is not. The algorithm has essentially 3 steps -- 1. apply hadamard to all bits (n bits to |0> and 1 bit to |1>). 2. Apply U_f, 3. Apply H again to all n bits. This time we don't apply the H gate to the last bit (which was initialized to |1>). So the y in equation 10, is just a variable that is similar to |x> in equation 7. So if we have 3 input bits 001, then we apply H to 001 and H to 1(this bit extra), then apply U_f, then apply H again to first 3 bits. so x.y is that pairwise product. Is it clearer now?