With = {a, b}, define L1 to be the language of the regular expression b(ab) (a + ) and L2 to be the language of strings that start with b and have an even number of as. Design DFAs M1 and M2 such that L(M1) = L1 and L(M2) = L2, with just a few appropriate strategy comments. Then expressly use the Cartesian Product construction to design a DFA M3 such that L(M3) = L(M1) L(M2). (You might fear that M3 would have a lot of states, but since this is the case and both L1 and L2 have dead conditions, you should note that you can economize quite a bit.( empty set of strings)