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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.


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

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  • $\begingroup$ 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