# Doubt in Regular expression

53 views

how to prove

$(x^{*}y)^{*}x^{*}y = x^{*}(yx^{*})^{*}y$

It contains extra y.

x*y(yx*)*y

1

We already know that $(PQ)^*P=P(QP)^*$

so take RHS  $x^* (yx^*)^* y=x^* y(x^* y)^*$

$=(x^*y)^+= LHS$

0

1
In $x^* (\color {green} y \color{red}{ x^*})^* \color{green} y$

P is y and Q is $x$*  then it becomes $x^* (P Q)^* P= x^* P(QP)^*= x^* y(x^* y)^*$
0

Thank you @Verma Ashish

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Tick is correct answer but how do we know which way to take common because it affect the language , is regular expression written in red is correct ?
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