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We are interested to determine if they are or not equivalent. We construct a table with elements of form (q, q ), where q ∈ Q and q ∈ Q . Beginning with the second column of the table, we associate a column to each letter of the alphabet Σ. If the first element of the ith row is (q, q ) then at the cross of ith row and the column associated to letter a will be the pair elem δ(q, a) , elem δ (q , a) . a ... ... (q, q ) elem δ(q, a) , elem δ (q , a) ... 2. Finite automata and regular languages 35 In the first column of the first row we put (q0 , q0 ) and complete the first row using the above method.

Qn−2 → an−1 qn−1 , qn−1 → an were used, which by definition means that in DFA A there is a walk a an−1 a a a n 3 2 1 qn , · · · −→ qn−1 −→ q2 −→ q1 −→ q0 −→ and since qn is a final state, u ∈ L(A) \ {ε} . If the DFA accepts also the empty word ε, then in the above grammar we introduce a new start symbol q0 instead of q0 , consider the new production q0 → ε and for each production q0 → α introduce also q0 → α. 13 Let A = ({q0 , q1 , q2 }, {a, b}, E, {q0 }, {q2 }) be a DFA, where E = (q0 , a, q0 ), (q0 , b, q1 ), (q1 , b, q2 ), (q2 , a, q2 ) .

12). 2. 13. • If (p, a, q) ∈ E for p, q ∈ Q and a ∈ Σ, then put production p → aq in P . • If (p, a, q) ∈ E and q ∈ F , then put also production p → a in P . Prove that L(G) = L(A) \ {ε}. Let u = a1 a2 . . an ∈ L(A) and u = ε. Thus, since A accepts word u, there is a walk an−1 a1 a2 a3 an q0 −→ q1 −→ q2 −→ · · · −→ qn−1 −→ q n , qn ∈ F . Then there are in P the productions q0 → a1 q1 , q1 → a2 q2 , . . , qn−2 → an−1 qn−1 , qn−1 → an (in the right-hand side of the last production qn does not occur, because qn ∈ F ), so there is the derivation q0 =⇒ a1 q1 =⇒ a1 a2 q2 =⇒ .

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