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Can anyone help me figure out the error in my approach to this problem from Sipser 1.18 (1.6f)? Write a regular expression for the language L = {w | w does not contain 110}

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So, the answer I get is: $(0 \cup 10)^* (1 \cup 111^* \cup \epsilon)$

And the answer given is: $(0 \cup (10)^*)^*1^*$

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2 Answers 2

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Note that $\mathcal{L}(1\cup 111^*\cup \varepsilon) = \mathcal{L}(1^*)$.

And in the general case, $\mathcal{L}((e\cup f)^*) = \mathcal{L}((e\cup f^*)^*)$.

So your regular expression is correct.

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To elaborate on Nathaniel's answer:

$1 \cup 111* \cup\space \epsilon \\ = 1(\epsilon \cup 11^*)\cup \epsilon \\ = 1(\epsilon + 1^+) \cup \epsilon \\ = 1(1^*) \cup \epsilon\\ = 1^+ \cup \epsilon \\ = 1^*$

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