State Machine - Last 4 Digits of Phone #

The last 4 digits of my phone number is 8 8 5 3. So for the state graph, S0 is 8 with Qa = 0 and Qb = 0, S1 is 8 with Qa = 0 and Qb = 1, S2 is 5 with Qa = 1 and Qb = 0, and S3 is 3 with Qa = 1 and Qb = 1. In binary, 8 is 1000, 5 is 0101, and 3 is 0011. For S0, C3 = 1, C2 = 0, C1 = 0, and C0 = 0. For S1, C3 = 1, C2 = 0, C1 = 0, and C0 = 0. For S2, C3 = 0, C2 = 1, C1 = 0, and C0 = 0. For S3, C3 = 0, C2 = 0, C1 = 1, and C0 = 1. I took all of this information and drew a state graph with it, shown below, sort of describing the process of the state machine. Then, I compiled the information into a table with the inputs and outputs. From this table, I got the unsimplified equations for Da, Db (the flip flop inputs), C3, C2, C1, and C0. I then used boolean logic to simplify these into the equations below. From then, I recreated the equations using AND and OR gates in multisim, along with two D flip flops and a 7-Segment decoder. A XOR gate is the same as two ANDs connected to an OR, making it a substitute in the circuit.