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11). 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 =⇒ .

Conversely, let L = ∅ and let u be the shortest word in L. We show that |u| < n. If |u| ≥ n, then we apply the pumping lemma, and give the decomposition u = xyz, |y| > 1 and xz ∈ L. This is a contradiction, because |xz| < |u| and u is the shortest word in L. Therefore |u| < n. 17 There exists an algorithm that can decide if a regular language is or not empty. Proof Assume that L = L(A), where A = (Q, Σ, E, {q0 }, F ) is a DFA. 15 language L is not empty if and only if it contains a word shorter than n, where n is the number of states of automaton A.

Ak z = ak+1 ak+2 . . am . This decomposition immediately yields to |xy| ≤ n and |y| ≥ 1. We will prove that xy i z ∈ L for any i. Because u = xyz ∈ L, there exists an walk x y z q0 −→ qj −→ qk −→ qm , qm ∈ F, and because of qj = qk , this may be written also as x y z q0 −→ qj −→ qj −→ qm , qm ∈ F . y From this walk qj −→ qj can be omitted or can be inserted many times. So, there are the following walks: x z q0 −→ qj −→ qm , qm ∈ F , x y y y z q0 −→ qj −→ qj −→ . . −→ qj −→ qm , qm ∈ F . 3 Pigeonhole principle: If we have to put more than k objects into k boxes, then at least one box will contain at least two objects.

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