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Self Doubt

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I tried to draw DFA for - string w such that there are even a's and every a is followed by at least one b.

First I took two dfa's one for even no of a's and other for strings where every a is followed by atleast one b and took their intersection as answer. I ended up with a solution having 6 states.

But there is a solution online for the same problem with 5 states. So, DFA obtained via intersection of two DFA's doesn't ensure minimal DFA ?
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