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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.

  • 2
    $\begingroup$ 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
    Commented Jan 20, 2019 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. $\endgroup$ Commented Jan 21, 2019 at 18:57
  • $\begingroup$ @ArunKumarSingh, hint, both $L_1$ and $L_2$ are $\{a^t\mid t\ge 1\}$. $\endgroup$
    – John L.
    Commented Jan 21, 2019 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. $\endgroup$ Commented Jan 24, 2019 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 $\{a^t\mid t\ge1\}$). $\endgroup$
    – John L.
    Commented Jan 24, 2019 at 16:05


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