Deterministic Finite Automata ( DFA)

Question

follow method first draw dfa for ‘a’ always followed by ‘bb’ and reverse it

Comments

what's your question ?

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Same doubt 😆

Answer 1

You have a small misunderstanding.

~$(\forall x,P(x)) \neq (\forall x,$~$P(x)) $ instead

1. ~$(\forall x,P(x)) = (\exists   x,$~$P(x)) $
2. $(\forall x,$~$P(x)) =$~$(\exists   x,$$P(x)) $

Above 3 lines in English  :

Complement of $a$ always followed by $bb$ is not $a$ never followed by $bb$ instead

1. Complement of $a$ always followed by $bb$ is at least one $a$ is not followed by $bb$.
2. $a$ never followed by $bb$ is complement of at least one $a$ is followed by $bb$.

Now about the DFA,

Let $D$ is DFA for language $L$ then $\bar D$ is DFA for language $\bar L$ where 

$\bar D$ is obtained my making all Final states in $D$ Non Final states and all Non Final states in $D$ becomes Final states in $\bar D$ ( Reversing the DFA as you said )

This sould have resolved any confusion, still here are all four diagrams for reference :