Show that $R^{+} \equiv R \leftrightarrow L(RR) \subset L(R)$

sigma is any alphabet. R is a regular expression.

How can L(RR) even be a subset or equal to L(R)?


If $\varepsilon\in L(R)$, then $RR$ matches every string matched by $R$: you can match the first $R$ against $\varepsilon$ and the second $R$ against the rest of the string.

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