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I have a regex $R=\epsilon+1+(\epsilon+1)(\epsilon+1)^*(\epsilon+1)$ which has to be simplified by algebraic operations. As we can do,

$\epsilon+1+(\epsilon+1)(\epsilon+1)^*(\epsilon+1)=\epsilon+1+(\epsilon+1)^+=(\epsilon+1)^+$ but the answer is $1^*$.

Can anyone please help me to find this with step by step solution?

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No. I'm not going to give a step-by-step solution, but here's a hint or two: $(\epsilon + 1)$ can generate either nothing or a 1; $(\epsilon + 1)(\epsilon + 1)$ can generate 0, 1, or 2 copies of 1. Take it from there.

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  • $\begingroup$ I understand this. But since I am applying algebraic operations, so what will be the correct law for which I get $(\epsilon+1)^*=1^*$. Again as mentioned in question, can $(\epsilon+1)^+$ generate $1^*$? $\endgroup$
    – Manjoy Das
    Jun 7, 2021 at 19:23
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    $\begingroup$ @ManjoyDas you should go back to the definition of the Kleene star and the concatenation with the empty word, it will help understanding. $\endgroup$
    – Nathaniel
    Jun 7, 2021 at 19:58
  • $\begingroup$ $1^*$ consists of all strings of the form $1^n$, for $n=0, 1, 2, \dots$, as yopu probably know. Ask yourself, from $(\epsilon + 1)$ give you $n=0$?, can it give you $n=1$?, can multiple concatenations can give you $n=2$?, $n=3$?. $\endgroup$ Jun 7, 2021 at 22:09

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