My Explanation On The Automata Assignment

Here is my homework for Automata Theory.
Pardon me for not being able to use MSPaint efficiently. ^^

As far as I can remember, the homework was to convert an NFA diagram to a DFA diagram.

Here's the NFA diagram:


Using subset construction, I am able to determine the states and state transitions.

δ_D({q₀},a) = δ_N(q₀,a) = {q₁}

δ_D({q₀},b) = δ_N(q₀,b) = {q₂}

δ_D({q₁},a) = δ_N(q₁,a) = {q₃}

δ_D({q₁},b) = δ_N(q₁,b) = {}

δ_D({q₂},a) = δ_N(q₂,a) = {}

δ_D({q₂},b) = δ_N(q₂,b) = {q₄}

δ_D({q₃},a) = δ_N(q₃,a) = {q₃} U {q₆} = {q₃,q₆}

δ_D({q₃},b) = δ_N(q₃,b) = {q₃}

δ_D({q₄},a) = δ_N(q₄,a) = {q₄} U {q₆} = {q₄,q₆}

δ_D({q₄},b) = δ_N(q₄,b) = {q₄}

δ_D({q₃,q₆},a) = δ_N(q₃,a) ∪ δ_N(q₆,a) = {q₃,q₆} ∪ {} = {q₃,q₆}

δ_D({q₃,q₆},b) = δ_N(q₃,b) ∪ δ_N(q₆,b) = {q₃} ∪ {} = {q₃}

δ_D({q₄,q₆},a) = δ_N(q₄,a) ∪ δ_N(q₆,a) = {q₄,q₆} ∪ {} = {q₄,q₆}

δ_D({q₄,q₆},b) = δ_N(q₄,b) ∪ δ_N(q₆,b) = {q₄} ∪ {} = {q₄}

δ_D({},a) = {}, δ_D({},b) = {}

From the subsets above we can produce this DFA:


However, we can still simplify the DFA (above). I'm not sure about the rules in simplifying DFAs yet, so I hope this will suffice for now:


Please leave your comments if I'm wrong or if I had broken some rules.