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Regular expressions are used to describe languages, in this article we discuss regular expressions and show that every language can be described by a regular expression.

### Table of contents.

- Introduction.
- Regular Expressions.
- Equivalence of regular expressions and regular languages.
- Summary.
- References.

### Prerequisites.

## Introduction.

*Regular expressions* are a means to describe languages, they define a finite pattern of *strings* or *symbols*.

Each pattern will match a set of strings and therefore they serve as names for a set of strings.

## Regular Expressions.

To get a idea about what regular expressions involve, let's go through a few examples:

*Example 1:*

Consider the expression *(0 âˆª 1)01**.

The language that is described by the above expression is the set of all binary strings,

- that start with
*0*or*1*as indicated by*0 âˆª 1*, - for which the second symbol is
*0*, - that end with zero or more
*1s*as indicated by*1**.

A sample language is such as the following;

*{00, 001, 0011, 00111, ..., 10, 101, 1011, 10111, ...}*

*Example 2:*

The expression *1*01*01** defines the language:

*{w : w contains exactly two 0s}*

*Example 3:* The expression *1011 âˆª 0* describes the language *{1011, 0}*.

*Example 4:* The expression *((0 âˆª 1)(0 âˆª 1))** describes the language *{w : the length of w is even}*

*Example 5:* The expression *(0 âˆª 1)*1011(0 âˆª 1)** describes the language *{w : 1011 is a substring of w}*

*Example 6:* The expression *(0 âˆª 1)*0(0 âˆª 1)*0(0 âˆª 1)** describes the language *{w : w contains at least two 0s}*

**Definition 1:** Let Î£ be a non-empty alphabet.

*Ïµ*is a regular expression.*âˆ…*is a regular expression.- For each
*a âˆˆ Î£*,*a*is a regular expression. - If
*R1*and*R2*are regular expressions, then*R1 âˆª R2*is also a regular expression. - If
*R1*and*R2*are regular expressions, then*R1R2*is also a regular expression. - If
*R*, then*R**is also a regular expression.

From the above listing, *1-3* are regarded as the building blocks of regular expressions while *3-6* define rules used to combine regular expressions into new/larger regular expressions.

*An example:* We claim that *(0 âˆª 1)*101(0 âˆª 1)** is a regular expression where the alphabet Î£ is {0, 1}.

**Proof:** We have to show that the expression can be built using the rules defined above.

- By rule
*3*,*0*and*1*are regular expressions. - By rule
*4*,*0 âˆª 1*is a regular expression since both*0*and*1*are regular expressions. - By rule
*5*,*10*and*101*are both regular expressions. - By rule
*5*,*(0 âˆª 1)*101*is a regular expression since*(0âˆª1)**and*101*are also regular expressions. - By rule
*5*,*(0 âˆª 1)*101(0 âˆª 1)**is a regular expression since*(0âˆª1)*101*and*(0âˆª1)**are both regular expressions. - By rule
*6*,*(0 âˆª 1)**is a regular expression.

Now we define the language that is described by the above regular expressions.

*Ïµ*describes the language*{Ïµ}*.*âˆ…*describes the language*{âˆ…}*.- For each
*a âˆˆ Î£*,*a*describes the language*{a}*. - Let
*R1*and*R2*be regular expressions and*L1*and*L2*be the languages described by these expressions respectively. The regular expressions*R1 âˆª R2*describes the language*L1 âˆª L2*. - Also assume the above for the languages
*L1*and*L2*and the expressions*R1*and*R2*, The regular expression*R1R2*also describes the language*L1L2*. - Let
*R*be a regular expression and*L*be the language described by it. The regular expression*R**describes the language*L**.

*Example 1:*

Regular expression *0 âˆª Ïµ* describes the language *{0, Ïµ}* whereas the expression *1** describes the language *{Ïµ, 1, 11, 111, ..., }* and therefore, the regular expression *(0 âˆª Ïµ)1** describes the language *{0, 01, 011, 0111, ..., Ïµ, 1, 11, 111, ...}*.

*Example 2:* Regular expression *(0 âˆª Ïµ)(1 âˆª Ïµ)* describes the language *{01, 0, 1, Ïµ}*.

**Definition 2:** Let *R1* and *R2* be regular expressions and *L1* and *L2* be languages described by the expressions, then if *L1* = *L2*, that is, *R1* and *R1* describe the same language, then we can write *R1* = *R2*.

Therefore, even if *(0 âˆª Ïµ)1** and *01** âˆª *1** are different regular expressions, we can write,

*(0 âˆª Ïµ)1* = 01* âˆª 1**

because they describe the same language.

**Theorem:** Let *R1*, *R2*, *R3* be regular expressions. The following identities hold.

- R1âˆ… = âˆ…R1 = âˆ….
- R1Ïµ = ÏµR1 = R1.
- R1 âˆª âˆ… = âˆ… âˆª R1 = R1.
- R1 âˆª R1 = R1.
- R1 âˆª R2 = R2 âˆª R1.
- R1(R2 âˆª R3) = R1R2 âˆª R1R3.
- (R1 âˆª R2)R3 = R1R3 âˆª R2R3.
- R1(R2R3) = (R1R2)R3.
- âˆ…* = Ïµ.
- Ïµ* = Ïµ.
- (Ïµ âˆª R1)* = R1*.
- (Ïµ âˆª R1)(Ïµ âˆª R1)* = R1*.
- R1*(Ïµ âˆª R1) = (Ïµ âˆª R1)R1*=R1*.
- R1*R2 âˆª R2 = R1*R2.
- R1(R2R1)* = (R1R2)*R1.
- (R1 âˆª R2)*= (R1*R2)*R1* = (R2*R1)*R2*.

We shall not present proof for the above identities however we assume they are verified.

*For example:* We can verify the identity *(0 âˆª Ïµ)1* = 01* âˆª 1** as follows,

- By identity
*7*: (0 âˆª Ïµ)1* = 01* âˆª Ïµ1* - By identity
*2*: (0 âˆª Ïµ)1* = 01* âˆª 1*

## Equivalence of regular expressions and regular languages.

**Theorem:** Let *L* be a language, then *L* is regular iff there exists a regular expression describing *L*.

The proof of this theorem entails two parts:

- First we will prove that every regular expression describes a regular language.
- Second, we prove that every DFA
*M*can be converted to a regular expression describing a language*L(M)*.

## 1. Every regular expression describes a regular language

Let *R* be an arbitrary regular expression over the alphabet Î£.

We need to prove that the language that is described by *R* is a regular language.

We prove using induction on the structure of *R*, that is, by induction on the way *R* is built using the rules we defined earlier(definition 1).

**Base case 1:** Assuming *R = Ïµ*, then *R* describes the language *{Ïµ}* and to prove that it is regular we shall construct a NFA *M = (Q,Î£, Î´, q, F)* that accepts this language.

We obtain this *NFA* by defining

*Q = {q}*, *q* the start state, *F = {q}* and *Î´(q, a) = âˆ…* for all a âˆˆ Î£Ïµ

State diagram of *M*:

**Base case 2:** Assuming *R = âˆ…*, then *R* describes the language âˆ… and to prove that it is regular, we construct an NFA *M = (Q,Î£, Î´, q, F)* that accepts it. We obtain this *NFA* by defining *Q = {q}*, *q* the start state, *F = âˆ…*, and *Î´(q, a) = âˆ…* for all *a âˆˆ Î£Ïµ*.

The state diagram of *M*:

**Base case 3:** Let *a âˆˆ Î£*, assume *R = a*, then *R* describes the language *{a}* and to prove that it is regular we construct a NFA *M= (Q,Î£, Î´, q1, F)* that accepts it. The *NFA* is obtained by defining *Q = {q1, q2}*, *q1* is the start state, *F = {q2}*, and

*Î´(q1, a) = {q2}*,

*Î´(q1, b)* = *âˆ… for all b âˆˆ Î£Ïµ \ {a}*,

*Î´(q2, b)* = *âˆ… for all b âˆˆ Î£Ïµ*.

State diagram of *M* is shown below.

**Induction step 1 of base case 1:** Assuming *R = R1 âˆª R2* where *R1* and *R2* are regular expressions. Let *L1* and *L2* be regular languages described by *R1* and *R2* respectively, then *R* describes the language *L1 âˆª L2* which is also regular by *Theorem 1.1* in the prerequisite article.

**Induction step of base case 2:** Assuming that *R = R1R2* where *R1* and *R2* are regular expressions. Let *L1* and *L2* be two regular languages described by the regular expressions respectively, then *R* describes the language *L1L2* which by *Theorem 4* in the prerequisite article is regular.

**Induction step of base case 3:** Assuming that *R = (R1)** where *R1* is regular expression. Let *L1* be the language described by *R1* and assuming that *L1* is regular, then *R* describes the language *(L1)** which by *Theorem 5* in the prerequisite article is regular.

At this point we conclude the proof of the claim that *every regular expression describes a regular language*.

*An example:*

Consider the regular expression: *(ab âˆª a)**, where *{a, b}* is the alphabet. We construct a *NFA* that accepts the language described by the regular expression to prove that the regular expression describes a regular language.

Lets first build a regular expression:

- First we take the regular expressions
*a*and*b*and combine them to a regular expression*ab*. - Next, we take the regular expressions
*ab*and*a*and combine them into regular expression*ab âˆª a*. - We take regular expression
*ab âˆª a*and transform it into the regular expression*(ab âˆª a)**.

Now to construct an *NFA*:

- We construct
*NFA M1*that accepts the language described by regular expression*a*:

- We construct
*NFA M2*that accepts the language described by regular expression*b*:

- Apply the construction from the proof of
*Theorem 4(prerequisite article)*to*M1*and*M2*which gives*NFA M3*that accepts the language described by the regular expression*ab*:

- Apply the construction from proof of
*Theorem 1.1:(prerequisite article)*to*M1*and*M3*which gives*NFA M4*that accepts the language described by regular expression*ab âˆª a*:

- Apply construction from proof of
*Theorem 5(prerequisite article)*to*M4*which gives*NFA M5*that accepts language described by regular expression*(ab âˆª a)**

## 2. Converting a DFA to a regular expression.

We will prove that every *DFA M* can be converted into a regular expression describing a language *L(M)*.

First we solve recurrence relations involving languages.

**Solving recurrence relations.**

Let *Î£* be an alphabet, *B* and *C* be known languages in *Î£** such that *Ïµ âˆ‰ B* and *L* be an unknown language such that *L = BL âˆª C*.

Let's try to express *L* in terms of *B* and *C* hence solving it.

To start consider an arbitrary string *u* in language *L*. Now to determine how *u* looks like. Since *u âˆˆ L* and *L = BL âˆª C*, we know *u* is a string in *BL âˆª C* and therefore we have the following possibilities:

*u*is an element of*C*.*u*is an element of*BL*, if this is the case, then there are strings*b âˆˆ B*and*v âˆˆ L*such that*u = bv*. Since*Ïµ âˆ‰ B*, we have*b â‰ Ïµ*and thus,*|v| < |u|*. Since*v*is a string in*L*which is equal to*BL âˆª C*,*v*is a string in*BL âˆª C*and we have the following possibilities for*v*.

*v*is an element of*C*, if this is the case,*u = bv*where*b âˆˆ B*and*v âˆˆ C*and therefore*u âˆˆ BC*.*v*is an element of*BL*, if this is the case, there are strings*bâ€² âˆˆ B*and*w âˆˆ L*such that*v = bâ€² w*. Since*Ïµ âˆ‰ B*we have*bâ€² â‰ Ïµ*, therefore,*|w|*<*|v|*. Since*w*is a string in*L*, there are two further possibilities.

***w*is an element of*C*, if this is the case,*u = bbâ€² w*, where*b*,*bbâ€² âˆˆ B*and*w âˆˆ C*, therefore*u âˆˆ BBC*.

***w*is an element of*BL*, if this is the case, there are strings*bâ€²â€² âˆˆ B*and*x âˆˆ L*such that*w = bâ€²â€²x*. Since*â‰ Ïµ B*we have*bâ€²â€² â‰ Ïµ*and therefore *|x| < |w|. Since*x*is a string in*L*,*x*is also a string in*BL âˆª C*and we have another two further possibilities for*x*.

****x*is an element in*C*, if this is the case,*u = bbâ€² bâ€²â€² x*where*b, bâ€², bâ€²â€² âˆˆ B*and*x âˆˆ C*and therefore*u âˆˆ BBBC*.

****x*is an element of*BL*and so on...

From the above we are convinced that any string *u* in *L* can be written as a concatenation of zero or more strings in *B* followed by one string in *C*.

**Lemma 1:** Let *Î£* be an alphabet and *B*, *C*, *L*, be languages in *Î£** such that *Ïµ â‰ B* and *L = BL âˆª C*, then *L = B*C*.

**Proof:**

We first show that *B*C âŠ† L*. Let *u* be an arbitrary string *B*C*. Then *u* is the concatenation of *k* strings of *B* for some *k â‰¥ 0* followed by one string of *C*.

*The base case*: k = 0, *u* is a string in *C*, therefore *u* is a string in *BL âˆª C*. Since *L = BL âˆª C*, *u* is a string in L.

*Inductive step:* *k â‰¥ 1*, we write *u = vwc* where *v* is a string in *B*, *w* is the concatenation of *k-1* strings of *B* and *c* is a string of *C*.

We define *y = wc* and observe that *y* is the concatenation of *k-1* strings of *B* followed by a single string of *C*.

By induction, string *y* is an element of *L*, therefore *u = vy* where *v* is a string in *B* and *y* a string in *L*.

This shows that *u* is a string in *BL* and therefore *u* is a string in *BL âˆª C* and since *L = BL âˆª C* if follows that *u* is a string in *L*.

We have completed the proof that *B*C âŠ† L*.

To show that *L âŠ† B*C*, let *u* be an arbitrary string in *L* and *l* its length. By induction we will prove that *u* is a string in *B*C*.

*Base case:* *l = 0*, *u = Ïµ* and since *u âˆˆ L* and *L = BL âˆª C*, *u* is a string in *BL âˆª C*. Since *Ïµ â‰ B*, *u* cannot be a string in *BL* and hence *u* must be a string in *C*.

Since *C âŠ† B*C*, *u* is a string in *B*C*.

*Induction step:* *l â‰¥ 1*, If *u* is a string in C, *u* is a string in *B*C* and we are done.

Assuming *u* is not a string in *C*. Since *u âˆˆ L* and *L = BL âˆª C*, *u* is a string in *BL*. Therefore there are strings *b âˆˆ B* and *v âˆˆ L* such that *u = bv*.

Since *Ïµ â‰ B*, the length of *b* is greater or equal to 1 and therefore the length of *v* is less than the length of *u*.

By induction, *v* is a string in *B*C* and hence *u = bv* where *b âˆˆ B* and *v âˆˆ B*C* which shows *u âˆˆ B(B*C)*. Since *B(B*C) âŠ† B*C*, it shows that *u âˆˆ B*C*.

This lemma holds for any language *B* that doesn't contains an empty string *Ïµ*.

**Converting DFA to regular expression.**

We use lemma 1 to prove that every DFA can be converted to a regular expression.

Let *M = (Q,Î£, Î´, q, F)* be an arbitrary deterministic finite automaton.

We will show that there exists a regular expression that describes a language *L(M)*.

For each state *r âˆˆ Q* we define:

*Lr={w âˆˆ Î£*: the path in the state diagram of M that starts in state r and corresponds to w ends in state F}.* That is, *Lr* is the language accepted by *M* , if *r* was the start state.

Now we show that *Lr* can be described by a regular expression, Since *L(M) = Lq*, it proves that *L(M)* can be described by a regular expression.

The idea is to set up equations for the languages *Lr* which we then solve using lemma 1.

We claim that

*${\mathrm{L}}_{\mathrm{r}}=\underset{\mathrm{a}\xe2\u02c6\u02c6\mathrm{\xce\pounds}}{\overset{}{\xe2\u2039\u0192}}\mathrm{a}.{\mathrm{L}}_{\mathrm{\xce\xb4}(\mathrm{r},\mathrm{a})\xc2}\xc2\mathrm{if}\xc2\mathrm{r}\xe2\u02c6\u2030\mathrm{F}$*

Let *w* be a string *Lr*, then the path *P* in the state diagram of *M* starting in state *r* and corresponding to *w* ends in state *F*.

Since *r âˆ‰ F*, this path contains at least one edge.

Let *râ€²* be the following state after the first state, then *râ€² = Î´(r, b)* for some symbol *b âˆˆ Î£*.

Therefore, *b* is the first symbol of *w*.

We write *w = bv* where *v* is the remaining part of *w*.

Then the path *Pâ€² = P \ {r}* in the state diagram *M* starting in state *râ€²* and corresponding to *v* ends in state *F*.

Thus, *v âˆˆ Lrâ€² = LÎ´(r, b)*.

Hence:

*$\mathrm{w}\xe2\u02c6\u02c6\mathrm{b}.{\mathrm{L}}_{\mathrm{\xce\xb4}\left(\mathrm{r},\mathrm{b}\right)\xc2}\xe2\u0160\u2020\underset{\mathrm{a}\xe2\u02c6\u02c6\mathrm{\xce\pounds}}{\overset{}{\xe2\u2039\u0192}}\mathrm{a}.{\mathrm{L}}_{\mathrm{\xce\xb4}\left(\mathrm{r},\mathrm{a}\right)\xc2}\xc2$*

Conversely, let *w* be a string in *${\xe2\u2039\u0192}_{\mathrm{a}\xe2\u02c6\u02c6\mathrm{\xce\pounds}}^{}\mathrm{a}.{\mathrm{L}}_{\mathrm{\xce\xb4}\left(\mathrm{r},\mathrm{a}\right)\xc2}$*, then there is a symbol *b âˆˆ Î£* and a string *v âˆˆ ${\mathrm{L}}_{\mathrm{\xce\xb4}\left(\mathrm{r},\mathrm{b}\right)\xc2}$* such that *w = bv*.

Let *Pâ€²* be a path in the state diagram of *M* starting in state *Î´(r, b)* and corresponding to *v*.

Since *v âˆˆ ${\mathrm{L}}_{\mathrm{\xce\xb4}\left(\mathrm{r},\mathrm{b}\right)\xc2}$*, this path ends in a state *F*.

Let *P* be the path in the state diagram of *M* starting in *r* and follows the edge to *Î´(r, b)*, then follows *Pâ€²*.

This path corresponds to *w* and end in a state *F* and therefore *w âˆˆ Lr*.

We complete the proof of the claim.

In a similar fashion we can prove that:

*${\mathrm{L}}_{\mathrm{r}}=\mathrm{\xcf\mu}\xe2\u02c6\xaa(\underset{\mathrm{a}\xe2\u02c6\u02c6\mathrm{\xce\pounds}}{\overset{}{\xe2\u2039\u0192}}\mathrm{a}.{\mathrm{L}}_{\mathrm{\xce\xb4}(\mathrm{r},\mathrm{a})\xc2})\xc2\mathrm{if}\xc2\mathrm{r}\xe2\u02c6\u2030\mathrm{F}$*

In this case we have a set of equations in the *unknowns* *Lr* for *r âˆˆ Q*.

The total number of equations is equal to the number of unknowns(size of *Q*).

To obtain the regular expression of *L(M) = Lq)* we solve the equations using lemma 1, for this we have to convince ourselves that these equations have a solution for any *DFA*.

**Lemma 2:** Let *n â‰¥ 1* be an integer and for *1 â‰¤ i â‰¤ n* and *1 â‰¤ j â‰¤ n*, let *${\mathrm{B}}_{\mathrm{ij}}$* and *${\mathrm{C}}_{\mathrm{i}}$* be regular expressions such that *Ïµ â‰ ${\mathrm{B}}_{\mathrm{ij}}$*. Let *L1, L2, ..., Ln* be languages satisfying:

*${\mathrm{L}}_{\mathrm{i}}=\left(\underset{\mathrm{j}=1}{\overset{\mathrm{n}}{\xe2\u2039\u0192}}{\mathrm{B}}_{\mathrm{ij}}{\mathrm{L}}_{\mathrm{j}}\right)\xe2\u02c6\xaa{\mathrm{C}}_{\mathrm{i}}\xc2\mathrm{for}\xc21\xe2\u2030\xa4\mathrm{i}\xe2\u2030\xa4\mathrm{n}$*

Then *L1* is expressed as a regular expression by involving regular expressions *${\mathrm{B}}_{\mathrm{ij}}$ and ${\mathrm{C}}_{\mathrm{i}}$*.

**Proof:** We prove this lemma by an induction on *n*.

*Base case:* n = 1, therefore, we have

*L1 = ${\mathrm{B}}_{\mathrm{11}}$ ${\mathrm{L}}_{1}$ âˆª ${\mathrm{C}}_{1}$*

Since *Ïµ âˆ‰ ${\mathrm{B}}_{\mathrm{11}}$*, by lemma 1 that *L1 = ${{\mathrm{B}}^{*}}_{11}$${\mathrm{C}}_{\mathrm{1}}$*, this proves the base case.

*Induction step:* Let *n â‰¥ 2*, we assume that the lemma is true for *n-1*, we have:

*${\mathrm{L}}_{\mathrm{n}}=\left(\underset{\mathrm{j}=1}{\overset{\mathrm{n}}{\xe2\u2039\u0192}}{\mathrm{B}}_{\mathrm{nj}}{\mathrm{L}}_{\mathrm{j}}\right)\xe2\u02c6\xaa{\mathrm{C}}_{\mathrm{n}}$*

= *${\mathrm{B}}_{\mathrm{nn}}{\mathrm{L}}_{\mathrm{n}}\xe2\u02c6\xaa\left(\underset{\mathrm{j}=1}{\overset{\mathrm{n}-1}{\xe2\u2039\u0192}}{\mathrm{B}}_{\mathrm{nj}}{\mathrm{L}}_{\mathrm{j}}\right)\xe2\u02c6\xaa{\mathrm{C}}_{\mathrm{n}}$*

Since *Ïµ âˆ‰ ${\mathrm{B}}_{\mathrm{nn}}$*, then from lemma 1:

*${\mathrm{L}}_{\mathrm{n}}={\mathrm{B}}_{\mathrm{nn}}^{*}\left(\left(\underset{\mathrm{j}=1}{\overset{\mathrm{n}-1}{\xe2\u2039\u0192}}{\mathrm{B}}_{\mathrm{nj}}{\mathrm{L}}_{\mathrm{j}}\right)\xe2\u02c6\xaa{\mathrm{C}}_{\mathrm{n}}\right)$*

= *${\mathrm{B}}_{\mathrm{nn}}^{*}\left(\underset{\mathrm{j}=1}{\overset{\mathrm{n}-1}{\xe2\u2039\u0192}}{\mathrm{B}}_{\mathrm{nj}}{\mathrm{L}}_{\mathrm{j}}\right)\xe2\u02c6\xaa{\mathrm{B}}_{\mathrm{nn}}^{*}{\mathrm{C}}_{\mathrm{n}}$*

= *$\left(\underset{\mathrm{j}=1}{\overset{\mathrm{n}-1}{\xe2\u2039\u0192}}{\mathrm{B}}_{\mathrm{nn}}^{*}{\mathrm{B}}_{\mathrm{nj}}{\mathrm{L}}_{\mathrm{j}}\right)\xe2\u02c6\xaa{\mathrm{B}}_{\mathrm{nn}}^{*}{\mathrm{C}}_{\mathrm{n}}$*

By substituting the equation for *Ln*, into equations for *Li*, *1 â‰¤ i â‰¤ n âˆ’ 1* we obtain the equation.

*${\mathrm{L}}_{\mathrm{i}}=\left(\underset{\mathrm{j}=1}{\overset{\mathrm{n}}{\xe2\u2039\u0192}}{\mathrm{B}}_{\mathrm{ij}}{\mathrm{L}}_{\mathrm{j}}\right)\xe2\u02c6\xaa{\mathrm{C}}_{\mathrm{i}}$*

= *${\mathrm{B}}_{\mathrm{in}}{\mathrm{L}}_{\mathrm{n}}\xe2\u02c6\xaa\left(\underset{\mathrm{j}=1}{\overset{\mathrm{n}-1}{\xe2\u2039\u0192}}{\mathrm{B}}_{\mathrm{ij}}{\mathrm{L}}_{\mathrm{j}}\right)\xe2\u02c6\xaa{\mathrm{C}}_{\mathrm{i}}$*

= *$\left(\underset{\mathrm{j}=1}{\overset{\mathrm{n}-1}{\xe2\u2039\u0192}}({\mathrm{B}}_{\mathrm{in}}{\mathrm{B}}_{\mathrm{nn}}^{*}{\mathrm{B}}_{\mathrm{nj}}\xe2\u02c6\xaa{\mathrm{B}}_{\mathrm{ij}}){\mathrm{L}}_{\mathrm{j}}\right)\xe2\u02c6\xaa{\mathrm{B}}_{\mathrm{in}}{\mathrm{B}}_{\mathrm{nn}}^{*}{\mathrm{C}}_{\mathrm{n}}\xe2\u02c6\xaa{\mathrm{C}}_{\mathrm{i}}$*

An therefore we have *n-1* equations in *L1, L2,...,Ln-1* and since *$\mathrm{\xcf\mu}\xe2\u02c6\u2030{\mathrm{B}}_{\mathrm{in}}{\mathrm{B}}_{\mathrm{nn}}^{*}{\mathrm{B}}_{\mathrm{nj}}\xe2\u02c6\xaa{\mathrm{B}}_{\mathrm{ij}}$*, it follows from the induction hypothesis that *L1* can be expressed as a regular expression only involving the regular expressions *${\mathrm{B}}_{\mathrm{ij}}$ and ${\mathrm{C}}_{\mathrm{i}}$*

## Summary.

Regular expressions are a means to describe languages, in this article, we proved that every regular expression describes a regular language and that every DFA can be converted to a regular expression that describes a language.