Closure Properties of Regular Sets

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Throughout the articles we have seen the finite state machines and regular languages and the regular sets. In this chapter, we will cover the basics of closure properties in regular sets which is the fundamental concept in finite automata and the study of formal languages.

Closure properties describe the behaviour of regular sets under various operations, ensuring that the application of these operations on regular sets results in another regular set.

Theorem on Closure Sets

Let us see the theorem for different closure properties on regular sets one by one.

Union Operation

Regular sets are closed under the union operation. This means if we have two regular sets L₁ and L₂, their union L₁ ∪ L₂ is also a regular set.

Proof − Let L₁ and L₂ be regular languages over an alphabet Σ. There exist two finite automata M₁ and M₂ such that

$$\mathrm{L_{1} \:=\: L(M_{1}) \:\:and \:\:L_{2} \: =\: L(M_{2})}$$

Define a new finite automaton M as follows −

• States of M are Q = Q₁ ∪ Q₂ ∪ {q₀}, where q₀ is a new start state not in Q₁ or Q₂.
• The start state is q₀.
• The final states are F = F₁ ∪ F₂
• The transition function δ includes −
  • δ(q₀, ) = {q₀₁, q₀₂}
  • Transitions of M₁ for states in Q₁
  • Transitions of M₂ for states in Q₂

Thus, any string accepted by M₁ or M₂ is accepted by M, proving that L₁ ∪ L₂ is regular.

Example

Consider we have two machines M₁ and M₂ accepting languages L₁ and L₂ respectively.

$$\mathrm{L_{1} \:=\: a_{*}\: (a\: +\: b)\: b^{*}}$$

$$\mathrm{L_{2} \:=\: ab(a \:+\: b)^{*}}$$

The machines will be like −

And,

After applying union, the machine will be like −

Complement Operation

The complement of a regular set is also a regular set. This property indicates that if L is a regular set, then its complement L^C is also regular.

Proof − If L is a regular set, there exists a finite automaton M that accepts L. Construct a new finite automaton M' with the same states and transitions as M, but with the final and non-final states swapped. The new automaton M' accepts the complement of L.

Let us see an example for language, L = ab(a + b)^*

Its complement will be the same machine but non-final will be final and final will be non-final.

Intersection Operation

Regular sets are closed under the intersection operation. This means if we have two regular sets L₁ and L₂, their intersection L₁ ∩ L₂ is also a regular set.

Proof − From De Morgans laws, we know that L₁ ∩ L₂ = $\mathrm{(L_1^C \cup L_2^C)}$. If L₁ and L₂ are regular, their complements L₁^C and L₂^C are also regular. The union of regular sets is regular, and the complement of a regular set is regular. Therefore, L₁ ∩ L₂ is regular.

Cross Product of a Regular Set

The cross product of a regular set is also a regular set. This property indicates that combination of two regular sets will fall under regular.

Proof − Let D₁ = {Q₁, Σ, δ₁, q₀₁, F₁} and D₂ = {Q₂, Σ, δ₂, q₀₂, F₂} be two DFA accepting two RE L₁ and L₂ respectively. Let us construct a new FA, D as follows.

$$\mathrm{D \:=\: \{Q,\: \Sigma,\: \delta,\: q_{0},\: F\}}$$

Where,

$$\mathrm{Q \:=\: Q_{1}\: \times\: Q_{2}}$$

$$\mathrm{\delta((S_{1},\: S_{2}),\: i/p) \:=\: (\delta_{1}(S_{1}, \:i/p),\: \delta_{2}(S_{2},\: i/p))\: \text{ for all }\: S_{1}\: \in \:Q_{1},\: S_{2}\: \in \:Q_{2}\: \text{ and } \:i/p \:\in \: \Sigma}$$

$$\mathrm{q_{0} \:=\: (q_{01},\: q_{02})}$$

$$\mathrm{F \:=\: F_{1}\: \times \: F_{2}}$$

So, D is a DFA. Thus, DFA are closed under cross product.

Conclusion

The closure properties of regular sets ensure that regular languages are closed under various operations such as union, intersection, complement, etc. These properties are foundational in automata theory in the context of regular languages.