3'3 Closure Properties of Regular Languages

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3.3 Closure Properties of Regular Languages
Given regular languages L 1 and L 2, we will show
that L 1? L 2, L 1? L 2, the complement of L
1and L 1 \ L 2 are regular. Sometimes, we can
apply the closure properties to show some sets
are not regular.
Theorem 3 Regular languages are closed under
union operation and Kleene star operation.
Proof
If L 1 and L 2 are regular, then there are
regular expressions r 1 and r 2 denoting the
languages L 1 and L 2 ,respectively.
( r 1 r 2) and (r 1) are regular expressions
denoting the languages L 1 ? L 2 and L 1.
Therefore, L 1 ? L 2 and L 1 are regular.
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Theorem 4 Regular languages are closed under
complement operation and intersection operation.
Proof
Suppose that L 1 and L 2 are regular over an
alphabet ?.
There is a DFA M(Q, ?, ?, q 0, F) accepting L 1.
Design a DFA M (Q, ?, ?, q 0, F), where F
Q \ F.
Now, we have that L(M) ? \ L 1. Hence, the
complement of L 1 is regular.
Let L 1 ? \ L 1 and L 2 ? \ L 2. The
complement of L 1 ? L 2 is regular and is equal
to L 1 ? L 2 . Hence L 1 ? L 2 is
regular.
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The proof of regular languages are closed under
intersection operation can be done by construct a
new DFA to accept the intersection. The
construction is as follows.
Suppose that L 1 and L 2 are regular languages
accepted by DFAs M 1(Q 1, ?, ? 1, q 0 1, F 1)
and M 2 (Q 2, ?, ? 2, q 0 2, F 2) such that L(M
1) L 1 and L(M 2 ) L 2.
Construct a DFA M (Q, ?, ?, q 0, F ), where Q
Q 1 ? Q 2,
q 0 (q 0 1, q 0 2), F F 1 ? F 2, and
?((p i, q s), a) (? 1(p i, a), ? 2(q s, a)).
It is not difficult to show that L(M) L 1 ? L
2.
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Example 9 Let L ? ? 0, 1 the number of
0s in ? is equal to the number of 1s in ?.
Show that L is not regular.
Solution
Assume that L were regular.
Let L L ? L(0 1 ).
L 0 i1 i i gt 0 . Since L and L (0 1 )
are regular, we have that L is regular.
By example 6 in section 3.2 we know that L is
not regular , a contradiction. Therefore, L is
not regular.
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Example 10 Let L 0 k k is a composite
integer. Show that L is not regular.
Solution
Let L 0 k k is a prime integer. Then L
0 \ L \ 0.
Assume that L were regular. Let n be the
constant in the pumping lemma.
Choose ? 0 s, where s p n, p n is the nth
prime number .
Let ? xyz, where xy ? n and y ? 1. Hence, x
0 x and y 0 y.
Consider the case that i p n 1, xy i z p
n p n y p n (y1), not a prime number.
Hence, contradict to that L is regular.
Therefore, L is not regular.