Transcript Closure Properties of Regular Sets

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.
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
Hence
1’  L 2’ is regular and is equal to L 1  L 2 .
L 1  L 2 is regular.
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 DFA’s
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.
Example 9: Let L = {  {0, 1}* | the number of 0’s in  is
equal to the number of 1’s in }. Show that L is not regular.
Solution :
Assume that L were regular.
Let L’ = L  L(0 + 1 +).
L’ = { 0 i1 i | i > 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.
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 (|y|+1), not a prime number.
Hence, contradict to that L’ is regular.
Therefore, L is not regular.