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Is it true to say that $\Sigma^* \cdot$ {$a^nb^n: n>=0$} = $\Sigma^*$

Becuase if we take $\Sigma^*$ and concatenate it to {$a^nb^n: n>=0$} we don't get any "new" words than those we had in $\Sigma^*$ in the first place.

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    $\begingroup$ yes, it is as $n \geq 0$. $\endgroup$ – OmG Dec 15 '18 at 15:29
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Yes if $\Sigma \supseteq \{a, b\}$. To show the equality, let's show one is a subset of the other, and vice versa.

  • $\Sigma^\ast \cdot \left\{ a^n b^n \middle| n \geq 0 \right\} \subseteq \Sigma^\ast$ holds, because for any word $w \in \Sigma^\ast \cdot \left\{ a^n b^n \middle| n \geq 0 \right\}$, we know $w = u a^n b^n$ for $u \in \Sigma^\ast$, therefore $w \in \Sigma^\ast$ since $\Sigma \supseteq \{a, b\}$.
  • $\Sigma^\ast \cdot \left\{ a^n b^n \middle| n \geq 0 \right\} \supseteq \Sigma^\ast$ holds, because

    $$ \Sigma^\ast \cdot \left\{ a^n b^n \middle| n \geq 0 \right\} \supseteq \Sigma^\ast \cdot \{ a^0 b^0 \} = \Sigma^\ast \cdot \{ \varepsilon \} = \Sigma^\ast $$

    where $\varepsilon$ is the empty word.

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