This document discusses properties of regular languages and proofs that regular languages are closed under various operations. It begins with an overview of operations like concatenation, union, star, complement, intersection, and reversal. It then provides examples of constructing finite state automata for each operation on sample regular languages. The key points made are that regular languages are closed under these operations and finite state automata can be constructed to recognize the resulting languages.

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Properties of Regular Languages
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3. © 2020 Wael Badawy

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1L 2L
21LLConcatenation:
*
1LStar:
21 LL ÈUnion:
Are regular
Languages
For regular languages and
we will prove that:
1L
21 LL Ç
Complement:
Intersection:
R
L1
Reversal:

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We say Regular languages are closed under
21LLConcatenation:
Star:
21 LL ÈUnion:
1L
21 LL Ç
Complement:
Intersection:
Reversal:
*
1L
R
L1

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a
b
b
a
NFA
Equivalent
NFA
e
e
a
b
b
a
A useful transformation: use one accept state
2 accept states
1 accept state

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NFA
Equivalent NFA
Single
accepting
state
e
e
e
In General

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NFA without accepting state
Add an accepting state
without transitions
Extreme case

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1LRegular language
( ) 11 LML =
1M
Single accepting state
NFA 2M
2L
Single accepting state
( ) 22 LML =
Regular language
NFA
Take two languages

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}{1 baL n
=
a
b
1M
{ }baL =2
ab
2M
0³n
Example

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Union
NFA for
11
1M
2M
21 LL È
e
21 LLw ÈÎ 21 or LwLw ÎÎ
e

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a
b
ab
}{1 baL n
=
}{2 baL =
}{}{21 babaLL n
È=ÈNFA for
Example
e
e

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Concatenation
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21LL
1M 2M
e
21LLwÎ
NFA for
221121 and: LwLwwww ÎÎ=
change to
regular state

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NFA for
14
a
b ab
}{1 baL n
=
}{2 baL =
}{}}{{21 bbaababaLL nn
==
e
Example

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Star Operation
15
*
L
Lwwwww ik Î= :21 !
LML =)(
e
e
NFA for
*
LwÎ
or e=w
M
change to
regular state

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**
1 }{ baL n
=
a
b
}{1 baL n
=
e
e
Example
NFA for

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Reverse
17
R
L
M
NFA for
M¢
1. Reverse all transitions
2. Make the initial state accept state
and the accept state initial state
LML =)( R
LML =¢)(

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}{1 baL n
=
a
b
1M
}{1
nR
baL =
a
b
1M¢
Example

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Complement
1. Take the DFA that accepts L
M
LML =)(
M¢
LML =¢)(
2. Make accept states regular
and vice-versa

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}{1 baL n
=
a
b
1M
ba,
ba,
}{},{ *
1 babaL n
-=
a
b
1M¢
ba,
ba,
Example

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NFAs cannot be used for complement
Make accept states regular
and vice-versa
{})( =ML
M M¢
)(}{)( MLML ¹=¢ e
**
},{)( baML =S=
NFA NFA
it is not the
complement

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Same example with DFAs
Make accept states regular
and vice-versa
{})( =ML
M M¢
)(},{)( *
MLbaML ==¢
**
},{)( baML =S=
DFA DFA
ba, ba,
it is the
complement

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Intersection
23
1L regular
2L regular
we show 21 LL Ç
regular

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DeMorgan’s Law: 2121 LLLL È=Ç
21 , LL regular, regular
21 , LL regular, regular
21 LL È regular
21 LL È regular
21 LL Ç regular

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Example
}{1 baL n
=
},{2 baabL =
regular
regular
}{21 abLL =Ç
regular

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1Lfor for 2LDFA
1M
DFA
2M
Construct a new DFA that accepts
Machine Machine
M 21 LL Ç
M simulates in parallel and1M 2M
Another Proof for Intersection Closure

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States in M
ji pq ,
1M 2MState in State in

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1M 2M
1q 2qa
transition
1p 2pa
transition
DFA DFA
11, pq a
New transition
MDFA
22, pq

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0q
initial state
0p
initial state
New initial state
00, pq
1M 2MDFA DFA
MDFA

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iq
accept state
jp
accept states
New accept states
ji pq ,
kp
ki pq ,
1M 2MDFA DFA
MDFA
Both constituents must be accepting states

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Example:
}{1 baL n
=
a
b
1M
0³n
}{2
m
abL =
b
b
2M
0q 1q 0p 1p
0³m
2q 2p
a
a
ba, ba,
ba,

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00, pq
DFA for intersection
}{}{}{)( ababbaML mn
=Ç=
10, pqa
21, pq
b
ab 11, pq
20, pq
a
12, pq
22, pq
b
ba,
a
b
ba,
b
a
M

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Construction procedure for intersection
1. Build Initial State
2. For each new state and for each symbol
add transition to either an existing state
or create a new state and point to it
3. Repeat step 2 until no new states
are added
4. Designate accept states

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00, pq
Automaton for intersection
}{}{}{ ababbaL mn
=Ç=
initial state

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00, pq
Automaton for intersection
}{}{}{ ababbaL mn
=Ç=
10, pqa
add transition and new state
for symbol a 1M 2M
0q 0qa
0p 1pa
00, pq a
M
10 , pq

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00, pq
Automaton for intersection
}{}{}{ ababbaL mn
=Ç=
10, pqa
21, pq
b
add transition and new state
for symbol b
1M 2M
0q 1qb 0p 2p
00, pq b
M
21, pq
b

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Automaton for intersection
}{}{}{ ababbaL mn
=Ç=
00, pq 10, pqa
21, pq
b
ab 11, pq
20, pq
a
12, pq
22, pq
b
ba,
a
b
ba,
b
a
Repeat until no new states can be added

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Automaton for intersection
}{}{}{ ababbaL mn
=Ç=
add Accept state
00, pq 10, pqa
21, pq
b
ab 11, pq
20, pq
a
12, pq
22, pq
b
ba,
a
b
ba,
b
a
accept state for
accept state for
1q 1M
1p 2M

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simulates in parallel and1M 2M
accepts string w if and only if:
accepts string w1M
and accepts string w2M
)()()( 21 MLMLML Ç=
Intersection DFA :M