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Am ∈ L and m ≥ n. Then, because the automaton accepts word u, there are states q0 , q1 , . . , qm and walk am−1 a1 a2 a3 am q0 −→ q1 −→ q2 −→ · · · −→ qm−1 −→ qm , qm ∈ F. Because the number of states is n and m ≥ n, by the pigeonhole principle3 states q0 , q1 , . . , qm can not all be distinct (see Fig. 17), there are at least two of them which are equal. Let qj = qk , where j < k and k is the least such index. Then j < k ≤ n. Decompose word u as: x = a1 a2 . . aj 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.

If L is regular, then there exists a DFA A = (Q, Σ, E, {q0 }, F ) for which L = L(A). Let q0 , q1 , . . , qn the k states of the automaton A. Dene languages Rij for all −1 ≤ k ≤ n and 0 ≤ i, j ≤ n. k Rij is the set of words, for which automaton A goes from state qi to state qj without using any state with index greater than k . Using transition graph we can say: a word k is in Rij , if from state qi we arrive to state qj following the edges of the graph, and concatenating the corresponding labels on edges we get exactly that word, not using k any state qk+1 , .

Q0 , a1 , q1 ) ∈ E, where q0 ∈ q 0 = I , thus, there is a walk ak−1 ak a1 a2 a3 q0 −→ q1 −→ q2 −→ · · · −→ qk−1 −→ qk , q0 ∈ I, qk ∈ F, so L(A) ⊆ L(A). In constructing DFA we can use the corresponding transition function δ :     δ(q, a) = δ(q, a) , ∀q ∈ Q, ∀a ∈ Σ.   q∈q 1. 5. The equivalent DFA with NFA A in Fig. 3. The empty set was excluded from the states, so we used here ∅ instead of {∅}. 10 to transform NFA A in Fig. 3. Introduce the following notation for the states of the DFA: S0 := {q0 , q1 }, S4 := {q0 , q2 }, S1 := {q0 }, S5 := {q1 , q2 }, S2 := {q1 }, S6 := {q0 , q1 , q2 } , S3 := {q2 }, where S0 is the initial state.

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