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How these two languages be regular.If there is comparison between m and n since (n < m) is the condition to be satisfied.

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marked as duplicate by Evil, Discrete lizard, Raphael Jan 20 at 17:56

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    $\begingroup$ Welcome to Computer Science! We discourage posts that simply state a problem out of context, and expect the community to solve it. Assuming you tried to solve it yourself and got stuck, it may be helpful if you wrote your thoughts and what you could not figure out. It will definitely draw more answers to your post. Until then, the question will be voted to be closed / downvoted. You may also want to check out these hints, or use the search engine of this site to find similar questions that were already answered. $\endgroup$ – Raphael Jan 20 at 17:55
  • $\begingroup$ This is not duplicate question i didn't find this question any where earlier. If there is question with explaination please provide the link. $\endgroup$ – Arun Kumar Singh Jan 21 at 18:57
  • $\begingroup$ @ArunKumarSingh, hint, both $L_1$ and $L_2$ are $\{a^t\mid t\ge 1\}$. $\endgroup$ – Apass.Jack Jan 21 at 22:51
  • $\begingroup$ @Apass.Jack we know that even { a^p | p is prime } is not regular language although it is also in the form {a^t | t>= 1} as we can not determine wether p is prime or not so similar condition is also here we can not differentiate in powers of "a" wether "m" is greater than "n" or not. $\endgroup$ – Arun Kumar Singh Jan 24 at 15:51
  • $\begingroup$ @ArunKumarSingh I am not saying $L_1$ is a subset of $\{a^t\mid t\ge1\}$. $L_1$ is the same as $\{a^t\mid t\ge1\}$. That is probably what tricked you. $L_2$ is $\{a^t\mid t\ge2\}$ (I was wrong saying "... and $L_2$ are $\{a^t\mid t\ge1\}$). $\endgroup$ – Apass.Jack Jan 24 at 16:05