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I know why A is irregular by Closure properties of irregular language. I also know the complement of $ \{ 0^n 1^n | n \in \mathbb{N}\}$ is $A = \{ 0^i 1^j| i \neq j\} \cup (0 \cup1)^*(1)(0 \cup1)^*0(0 \cup1)^*$, but I don't see how $A^* = \sum^*$. Any help would be appreciated!

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  • $\begingroup$ You did a fine job presenting "your own formulas" using MathJax: please make the problem statement (?) a block quote, consider presenting it using MathJax (searchable) instead of hyperlinking a pixel raster. $\endgroup$
    – greybeard
    Jun 11 '20 at 9:41
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Since $0 \in A$ and $1 \in A$, any word $w \in \Sigma^*$ can be written has a concatenation of $n=|w|$ words $w_1, \dots, w_n$ such that each $w_i \in A$ (showing that $w \in A^*$). In order to do so it suffices to pick $w_i$ as the $i$-th character of $w$, i.e., $w_i$ is either $0$ or $1$.

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  • $\begingroup$ Oh yea right! I missed the fact that 0 in A and 1 in A. thanks $\endgroup$ Jun 10 '20 at 16:56

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