By Ivanyi A. (ed.)

Ivanyi A. (ed.) Algorithms of informatics, vol.1.. foundations (2007)(ISBN 9638759615)

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**Example text**

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 . Therefore xy i z ∈ L for all i, and this proves the theorem.

Xn , where for any i = 1, 2, . . , n expression xi is a regular expression representing language {wi }. This latter can be done by the following rule. If wi = ε, then xi = ε, else if wi = a1 a2 . . am , where m ≥ 1 depends on i, then xi = a1 a2 . . am , where the brackets are omitted. We prove the theorem of Kleene which refers to the relationship between regular languages and regular expression. 20 (Kleene's theorem). Language L ⊆ Σ∗ is regular if and only if there exists a regular expression over Σ representing language L.

Then, by the denition of the transitions of NFA A there exists a walk a a an−1 a a 1 2 3 n S −→ A1 −→ A2 −→ · · · −→ An−1 −→ Z, Z ∈ F. Thus, u ∈ L(A). If ε ∈ L(G), there is production S → ε, but in this case the initial state is also a nal one, so ε ∈ L(A). Therefore, L(G) ⊆ L(A). Let now u = a1 a2 . . an ∈ L(A). Then there exists a walk a a an−1 a a 1 2 3 n S −→ A1 −→ A2 −→ · · · −→ An−1 −→ Z, Z ∈ F. If u is the empty word, then instead of Z we have in the above formula S , which also is a nal state.