1. Regular Expressions • Equivalence Of Reg. Exps. And Finite Automata
• Kleenex Algebra
• Languages defined by regular expressions are called REGULAR SETS.
004 Regular Expressions

2. Regular Expressions Over Alphabet 
1.  is a Reg. Exp.
|||| = 
2.  is a Reg. Exp.
|||| = {}
3. a  a is a Reg. Exp.
||a|| = {a}
4. , : Reg. Exp. 
(+), (), (*) : Reg. Exp.
||+|| = ||||||||,
|||| = ||||||||
||*|| = (||||)*
5. Nothing else is a Reg. Exp. Over .
004 Regular Expressions

3. From Reg. Exps. To -NFAs (1)
Algorithm REtoNFA
1a: If =, then
M accepts .
1b: If =, then
M accepts {}.
1c: If =a, then
Ma accepts {a}.
2a: If =, then s f sf a s f  s f s f M M
M

4. From Reg. Exps. To -NFAs (2)
2b: If =+, then
2c: If =*, then   s f M s f  s f  M
M+    f s s f M
M* 

5. From DFAs To Reg. Exps. Rijk={x* | (i) *(qi,x)=qj, Rijk=
Let M=({q1,..,qn},,,q1,F)
be a DFA.
For 1i,jk; 0kn, define
Rijk={x* | (i) *(qi,x)=qj,
(ii) x=yz & y+ & z+&
& *(qi,y)=qm  mk}.
Then
{a | (qi,a)=qj} if ij,
{a | (qi,a)=qj}{} if i=j,
and if k>0, then
Rijk=Rikk-1(Rkkk-1)*Rkjk-1Rjjk-1.
Rijk=
004 Regular Expressions

6. Regular Expressions = Finite Automata
THEOREM
L(DFA)=L(Reg Exp).
Proof: :
Reg.Exp., -NFA M [ L(M)=|||| ]
by Algorithm REtoNFA. :
DFA M, Reg.Exp. [ ||||=L(M) ]
i) Prove by Induction on k that
Rijk, Reg.Exp.ijk [ ||ijk||=Rijk ].
ii) L(M)=||1f1n+ 1f2n+…+ 1fmn||, where
F={qf1, …, qfm}.
004 Regular Expressions

7. Kleene Algebra A Complete Axiom System for Regular Expressions
Axioms for K=(,+,,0,1)
A1 a+(b+c)=(a+b)+c, a(bc)=(ab)c
A2 a+b=b+a
A3 a+a=a
A4 a+0=a, a1=1a=a
A5 a0=0a=0
A6 a(b+c)=ab+ac, (a+b)c=ac+bc
A7 1+aa*=a*, 1+a*a=a*
A8 abb  a*bb, and bab  ba*b,
where ab  a+b=b
for all a,b,c in .
( is omitted from xy.)
def
004 Regular Expressions

8. The Axiom System is Complete and Sound.
THEOREM All and only properties of reg. exps. over  can be derived from axioms A1 to A8.
Ex. (1) a*a*=a*, since
a*+a*a* = a*+a*(1+aa*) = a*+(a*+a*aa*) = (a*+a*)+a*aa*
= a*+a*aa* = a*(1+aa*) = a*a*.
(2) a* = 1+aa* = (1+1)+aa* = 1+(1+aa*) = 1+a*.
(3) ab & cd  a+b=b & c+d=d
 ac+bd = ac+(a+b)(c+d) = ac+(ac+ad+bc+bd) = (ac+ac)+ad+bc+bd = ac+ad+bc+bd =(a+b)(c+d) = bd  acbd. A7
A6,A4
   L A1
A4,A6 A7 A3 A3 A1 A7 A7
def of 
A6,A2,A1 A1 A3
A1,A6
def of 