a. Let A be a finite-state automaton with input alphabet
, and suppose
is the language accepted by A. The complement of
is the set of all strings over
that are not in
. Show that the complement of a regular language is regular by proving the following:
is the language accepted by a finite-state automaton A, then there is a finite-state automaton
that accepts the complement of
b. Show that the intersection of any two regular languages is regular as follows: First prove that if and are languages accepted by automata and , respectively, then there is an automaton A that accepts . Then use one of Dc Morgan’s laws for sets, the double complement law for sets, and the result of part (a) to prove that there is an automaton that accepts .
That the complement of a regular language is also a regular language by showing that there exists a finite-state automaton accepts the complement of .
A finite-state automaton is given that accepts the language which is defined on an alphabet . The complement of is a set of strings on that not belong to .
Kleene’s theorem for accepted language.
If any language is accepted by a finite-state automaton, then there exists a regular expression that defined the subjected language.
If any language can be expressed by a regular expression, then there exists a finite-state automaton that accepts the same language.
Because is accepted by , by the Kleene’s theorem there exists a regular expression that can be used to express
That the intersection of two regular languages is regular by proving the union of complements of the two languages is accepted by a finite-state automaton.
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