4 4: If L1 and L2 are regular languages, then L1/L2 is regu- lar: The family of regular languages is closed under right quotient with a regular language Proof: 1 Assume that L1 and L2 are regular, and let DFA M = (Q,Σ, δ, q0,F) accept L1
Very useful in studying the properties of one language by relating it to other ( better If L is regular, then there is a DFA M = (Q,Σ, δ, q0,F) such that L = L(M) Regular Languages are closed under intersection, i e , if L1 and L2 are regular then
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n > 0} Closure Properties A set is closed over an operation if L1, L2 ∈ class L1 op L2 = L3 ⇒ L3 Theorem 4 1 If L1 and L2 are regular languages, then L1 ∪ L2 L1 ∩L2 L1L2 ¯ L1 L ∗ To Use the Pumping Lemma to prove L is not
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If L1 ⊆ L2 and L2 is regular, then L1 must be regular There are subsets of a regular language which are not regular If L1 and L2 are nonregular, then L1 ∪ L2 must be nonregular
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1 what will happen when we perform operations on 3 how can we tell if a given language is regular or not? If L1 and L2 are regular languages, then so are
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that operation to any members of the set always yields a member of the That is, if L1 and L2 are regular languages, then each of L1 ∪ L2 , L1 L2 and L1 ∗
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If L1 and L2 are regular languages, then so are Liu L2, Lin L2, LjL2, L1, and Li To form the right quotient of Li with L2, we take all the strings in L1 that have a
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17 oct 2013 · (d) For any two languages L1 and L2, if L1 ∪ L2 is regular, then L1 and L2 are from X to Y Prove that Covers is reflexive but not symmetric
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all possible transitions on a for each state in S, then taking the set of states reachable If L1 and L2 are regular languages, is L1 ∪ L2? start ε ε Machine for L 1
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It is well known that the intersection of a context free language and a regular language is context free This theorem is used in several proofs that certain
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Closure under ?. 1. Page 2. Proposition 4. Regular Languages are closed under intersection i.e.
multiplication? subtraction? division? Closure of Regular Languages. Theorem 4.1 If L1 and L2 are regular languages then. L1. ? L2. L1. ?L2. L1L2. ¯. L1.
11 août 2000 FALSE. Let L1 = ?? and let L2 be any nonregular language over ?. (3) If L1L2 is regular and L1 is finite then ...
I'll fix them immediately. Thm. 4.1: If L1 and L2 are regular languages then so are L1 ?L2
If L? is regular and L2 is regular L1 must be regular. (c) If L is regular
Closure under ?. 1. Page 2. Proposition 4. Regular Languages are closed under intersection i.e.
multiplication? subtraction? division? Closure of Regular Languages. Theorem 4.1 If L1 and L2 are regular languages then. L1 ? L2. L1 ?L2. L1L2. ¯. L1.
Theorem 2.3.1 If L1 and L2 are regular languages then. L1. L2. L1L2. L*. 1. L1. L1. L2 are regular languages. Proof sketch. Union M1 = K1; ; 1; s1;
If L1 and L2 are regular languages so are L1 ? L2
L1 L2 = L1 ? L2 and regular languages are closed under intersection and complemen- tation. (B) If L ? ?? and L is finite
Thm 4 4: If L1 and L2 are regular languages then L1=L2 is regu-lar: The family of regular languages is closed under right quotient with a regular language Proof: 1 Assume that L1 and L2 are regular and let DFA M = (Q;?;–;q0;F) accept L1 2 We construct DFA Md = (Q;?;q0;Fc) as follows (a) For each qi 2 Q determine if there is a y 2
L1? L2is regular since regular languages are closed under complement ? L1? L2 is regular Proof via finite automata construction Let M? M?? and M be the formal definition of the finite automata that recognizes L1 L2 and L1? L2respectively M?=(Q? ? ?? q0? F?) M??=(Q???????? q0?? F??)
L1 and L2 are regular languages)9reg expr r1 and r2 s t L1 =L(r1)andL2=L(r2) r1 + r2 is r e denoting L1 [L2)closed under union r1r2 is r e denoting L1L2)closed under concatenation r 1 is r e denoting L 1)closed under star-closure 3
L1 and L2 are regular languages )9reg expr r andr s t L1 =L(r1) and L2=L(r2) r+r 2 is r e denoting L1[L2 )closed under union r1r2 is r e denoting L1L2 )closed under concatenation is r e denoting L 1 )closed under star-closurecomplementation:L1 is reg lang )9DFA M s t L1 = L(M)Construct M' s t
By de nition of regular languages we can say: If L1 and L2 are two regular languages then L1[L2 L1L2 L1 are regular 2 How about L1L2 L1 L2 ? also regular 3 Regular languages are closed under union concata-nation Kleene star set intersection set di erence etc 4 Given two FAs M1 and M2 can we construct new FAs to accept the the
If L and M are regular languages then so is L – M = strings in L but not M Proof: Let A and B be DFA’s whose languages are L and M respectively Construct C the product automaton of A and B Make the final states of C be the pairs where A-state is final but B-state is not
What if L1 is regular?
(b) Let L? = L1 ? L2. If L? is regular and L2 is regular, L1 must be regular. FALSE. We know that the regular languages are closed under intersection. But it is important to keep in mind that this closure lemma (as well as all the others we will prove) only says exactly what it says and no more.
Which subset of a regular language is regular?
Every subset of a regular language is regular. Let L? = L1 ? L2. If L? is regular and L2 is regular, L1 must be regular. If L is regular, then so is L? = {xy : x ? and y ? L}.
Is L? a concatenation of two regular languages?
(c) If L is regular, then so is L? = {xy: x ? L and y ? L}. TRUE. Proof: Saying that y ? L is equivalent to saying that y ? L. Since the regular languages are closed under complement, we know that L is also regular. L? is thus the concatenation of two regular languages. The regular languages are closed under concatenation. Thus L? must be regular.
Why is the intersection of L and LR regular?
is regular because the problem statement says so. LR is also regular because the regular languages are closed under reversal. The regular languages are closed under intersection. So the intersection of L and LR must be regular.
1 Closure Properties