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Jul 25, 2019 at 21:06 comment added John L. Here is another general construction. Let $\Sigma=\{a,b\}$. Let $L$ be a non-regular context-free language over $\Sigma$. Let $L_a = \{\epsilon\}\cup aL \cup b\Sigma^*$ and $L_b = \{\epsilon\}\cup a\Sigma^*\cup bL $. Then both $L_a$ and $L_b$ are non-regular context-free. $L_a \cup L_b = \Sigma^*$.
Jul 24, 2019 at 22:28 vote accept leonard9500
Jul 24, 2019 at 20:52 history edited John L. CC BY-SA 4.0
Added two exercises.
Jul 24, 2019 at 20:08 history answered John L. CC BY-SA 4.0