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Arden’s theorem helps in checking the equivalence of two regular expressions.

Let, P and Q be two regular expressions over the input set Σ. The regular expression R is given as follows −

**R=Q+RP**

This has a unique solution as **R=QP***.

Let, P and Q be the two regular expressions over the input string Σ.

If P does not contain ε then there exists R such that

**R= Q+RP-----------------------equation 1**

We will replace R by QP* in equation 1

Consider Right hand side (R.H.S) of equation 1

= Q+QP*P

=Q(ε+P*P)

=QP* since R*R=R* according to identity

Thus R=QP* is proved.

To prove that R=QP* is a unique solution, we will now replace left hand side (L.H.S) of eq 1 by Q+RP

Then it becomes, Q+RP

But again, R can be replaced by Q+RP

Therefore,

Q+RP = Q+(Q+RP)P

=Q+QP+PP^{2}

Again, replace R by Q+RP

=Q+QP+(Q+RP)P^{2}

=Q+QP+QP^{2}+RP^{3}

Thus, if we go on replacing R by Q+RP then we get,

Q+RP=Q+QP+QP^{2}+………+QP^{i}+RP^{i+1}

=Q(ε+P+P^{2}+…….+P^{i})+RP^{i+1}

From equation 1

**R = Q(ε+P+P2+…….+Pi)+RP ^{i+1} ---------------equation 2**

Where i>=0

Consider equation 2

R= Q(ε+P+P2+…….+Pi)+RP^{i+1}

R= QP*+RP^{i+1}

Let w be a string of length i

In RPi+1 has no string of less than i+1 length.

Hence,

- w is not in set RPi+1
- R and QP* represent the same set.

Hence, it is proved that

R=Q+RP has a unique solution

R=QP*

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