>>2476894

Oui

>>2476958

Statement −

Let P and Q be two regular expressions.

If P does not contain null string, then R = Q + RP has a unique solution that is R = QP*

Proof −

R = Q + (Q + RP)P [After putting the value R = Q + RP]

= Q + QP + RPP

When we put the value of R recursively again and again, we get the following equation −

R = Q + QP + QP2 + QP3…..

R = Q (ε + P + P2 + P3 + …. )

R = QP [As P represents (ε + P + P2 + P3 + ….) ]

>>2477019

:D