Given $\sum_{1}=\{a, b\}$ and $\sum_{2}=\{0, 1\}$, is it true to say that $b1\lambda0b \in (\sum_{1} \cup \sum_{2})^{+}$?

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Given $\sum_{1}=\{a, b\}$ and $\sum_{2}=\{0, 1\}$, is it true to say that $b1\lambda0b \in (\sum_{1} \cup \sum_{2})^{+}$?

My reasoning is that

$(\sum_{1} \cup \sum_{2})^{+} = (\sum_{1} \cup \sum_{2})(\sum_{1} \cup \sum_{2})^{*} = (\sum_{1} \cup \sum_{2})(\sum_{2}^{*} \cup \sum_{1}^{*})^{*}$,

and therefore

$b1\lambda0b \in (\sum_{1} \cup \sum_{2})^{+}$ must be true because $b \in \sum_{1}$, $1\lambda0 \in \sum_{2}^{*}$ and $b \in \sum_{1}^{*}$.

Is this correct?

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Your reasoning is correct.

Indeed, $\Sigma_1=\{a,b\}$ and $\Sigma_2=\{0,1\}$, so $\Sigma_1\cup \Sigma_2=\{a,b,0,1\}$, and thus $(\Sigma_1\cup\Sigma_2)^+$ contains any string $\neq\varepsilon$ made up of $a$'s, $b$'s, $0$'s and $1$'s.


Note. Typically, the notation $L^+$ is defined by $L^+=LL^*$. In other words, $L^+ = L^* \smallsetminus\{\varepsilon\}$.