Prove That the Family of Regular Languages Is Closed Under the Suff Operation,

Closure properties of Regular languages

Closure backdrop on regular languages are defined equally certain operations on regular linguistic communication which are guaranteed to produce regular language. Closure refers to some functioning on a language, resulting in a new language that is of aforementioned "type" as originally operated on i.eastward., regular.

Regular languages are closed under following operations.

Consider Fifty and M are regular languages:

1. Kleen Closure:
   RS is a regular expression whose language is L, M. R* is a regular expression whose language is L*.
2. Positive closure:
   RS is a regular expression whose language is L, M. is a regular expression whose language is .
3. Complement:
   The complement of a language L (with respect to an alphabet such that contains L) is –50. Since is surely regular, the complement of a regular language is ever regular.
4. Reverse Operator:
   Given linguistic communication L, is the fix of strings whose reversal is in L.
   Case: 50 = {0, 01, 100};
   ={0, 10, 001}.
   Proof: Let E be a regular expression for L. We prove how to contrary E, to provide a regular expression for .
5. Complement:
   The complement of a language L (with respect to an alphabet such that contains L) is –L. Since is surely regular, the complement of a regular language is always regular.
6. Union:
   Let L and M be the languages of regular expressions R and S, respectively.Then R+Due south is a regular expression whose language is(L U Thousand).
7. Intersection:
   Permit L and Thou exist the languages of regular expressions R and South, respectively and so information technology a regular expression whose language is 50 intersection Grand.
   proof: Allow 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 consisting of final states of both A and B.
8. Ready Difference operator:
   If L and M are regular languages, then so is L – One thousand = strings in L simply non 1000.

   Proof: Let A and B exist DFA'due south whose languages are L and Yard, respectively. Construct C, the production automaton of A and B brand the final states of C be the pairs, where A-state is terminal merely B-land is not.

9. Homomorphism:
   A homomorphism on an alphabet is a office that gives a cord for each symbol in that alphabet. Example: h(0) = ab; h(1) = . Extend to strings by h(a1…an) =h(a1)…h(an). Example: h(01010) = ababab.

   If L is a regular language, and h is a homomorphism on its alphabet, then h(L)= {h(due west) | w is in L} is also a regular language.
   Proof: Permit E be a regular expression for L. Apply h to each symbol in E. Language of resulting R, E is h(50).

10. Inverse Homomorphism : Let h be a homomorphism and 50 a linguistic communication whose alphabet is the output language of h. (50) = {w | h(w) is in Fifty}.

Note: In that location are few more properties similar symmetric difference operator, prefix operator, substitution which are closed under closure properties of regular language.

Decision Backdrop:
Approximately all the properties are decidable in case of finite automaton.

            (i)            Emptiness            (2)            Non-emptiness            (iii)            Finiteness            (4)            Infiniteness            (v)            Membership            (6)            Equality          

These are explained equally following below.

(i) Emptiness and Non-emptiness:

• Step-1: select the state that cannot exist reached from the initial states & delete them (remove unreachable states).
• Step ii: if the resulting machine contains at least one final states, so then the finite automata accepts the not-empty language.
• Footstep 3: if the resulting machine is free from final state, then finite automata accepts empty language.

  (two) Finiteness and Infiniteness:

  • Pace-1: select the country that cannot be reached from the initial state & delete them (remove unreachable states).
  • Step-two: select the land from which nosotros cannot reach the final state & delete them (remove dead states).
  • Pace-3: if the resulting machine contains loops or cycles then the finite automata accepts infinite language.
  • Footstep-iv: if the resulting machine exercise not contain loops or cycles and so the finite automata accepts infinite linguistic communication.

  (iii) Membership:
  Membership is a property to verify an arbitrary string is accepted by a finite automaton or non i.eastward. information technology is a member of the linguistic communication or not.

  Let M is a finite automata that accepts some strings over an alphabet, and let 'west' be any string defined over the alphabet, if there exist a transition path in 1000, which starts at initial country & ends in anyone of the terminal state, then string 'w' is a fellow member of M, otherwise 'w' is not a member of Chiliad.

  (iv) Equality:
  2 finite state automata M1 & M2 is said to be equal if and just if, they take the same language. Minimise the finite state automata and the minimal DFA will exist unique.

  
  

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Source: https://www.geeksforgeeks.org/closure-properties-of-regular-languages/

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