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LENGUAJES FORMALES Y AUTOMATA
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The regular expressions ( a^* b^* )^* and
( a b )^* generate the same regular language.
The class of regular languages is closed under intersection
If M is a finite automaton, then
L(M) is a regular language
The regular expressions ( a \cup b )^* and
(a^* b^* )^* generate two different regular languages
A DFA may have multiple initial states
Suppose that N_1 = (Q_1, \sum, \delta_1, q_1, F_1) recognizes the regular language
A_1, and
N_2 = (Q_2, \sum, \delta_2, q_2, F_2) recognizes the regular language
A_2. To prove that
A_1\circ A_2 is a regular language, we contruct the finite automata
N = (Q, \sum, \delta, q_0, F) where
Suppose that N_1 = (Q_1, \sum, \delta_1, q_1, F_1) recognizes the regular language
A_1. To prove that
A_1^* is a regular language, we contruct the finite automata
N = (Q, \sum, \delta, q_0, F) where
Suppose that N_1 = (Q_1, \sum, \delta_1, q_1, F_1) recognizes the regular language
A_1, and
N_2 = (Q_2, \sum, \delta_2, q_2, F_2) recognizes the regular language
A_2. To prove that
A_1\circ A_2 is a regular language, we contruct the finite automata
N = (Q, \sum, \delta, q, F) where
Suppose that N_1 = (Q_1, \sum, \delta_1, q_1, F_1) recognizes the regular language
A_1, and
N_2 = (Q_2, \sum, \delta_2, q_2, F_2) recognizes the regular language
A_2. To prove that
A_1\cup A_2 is a regular language, we contruct the finite automata
N = (Q, \sum, \delta, q_0, F) where
If a finite automaton accepts no strings, it still recognizes one language