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Converse of pumping lemma for regular expressions


Pumping lemma for simple finite regular languagesProofs using the regular pumping lemmaShow that the pumping lemmas for context-free and regular languages are equivalent for unary languagesProof that a language is not regular using pumping lemmaShowing that the pumping lemma cannot prove that some language is not regularWhat's wrong with my pumping lemma proof?How to show that a turtle graphic language is not regular?Prove this language is non-regularPumping-Lemma regular languages: Consider multiple cases?How to prove using pumping lemma that language generated by a(b*)c(d*)e is regular?













2












$begingroup$


I want to come up with a language that satisfies the pumping lemma while not being a regular expression.



I thought of $0^i1^j: i > j > 0 $. The pumping seems to work just fine, and this is not a regular expression, as it would not be possible to create an automata that recognizes it. This last part is where I'm having a bit of trouble, as I think I have the right idea, but can't formalize it. This language is just a regular expression $big(0^i : i in mathbbNbig)$ concatenated with something that is not a regular expression $big(0^i1^i: i in mathbbNbig)$.



Is my thought process correct? Can this be generalized? If $L$ is a RE and $L'$ is not, then $LL'$ is also not a RE? I believe so, because we know that if $L$ is a RE then so is $L^r$. And so we end up with $(L')^rL^r$, and we can use the contrapositive of the pumping lemma to prove this language is not regular. Any insight would be helpful!










share|cite|improve this question







New contributor



mathmathmath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$
















    2












    $begingroup$


    I want to come up with a language that satisfies the pumping lemma while not being a regular expression.



    I thought of $0^i1^j: i > j > 0 $. The pumping seems to work just fine, and this is not a regular expression, as it would not be possible to create an automata that recognizes it. This last part is where I'm having a bit of trouble, as I think I have the right idea, but can't formalize it. This language is just a regular expression $big(0^i : i in mathbbNbig)$ concatenated with something that is not a regular expression $big(0^i1^i: i in mathbbNbig)$.



    Is my thought process correct? Can this be generalized? If $L$ is a RE and $L'$ is not, then $LL'$ is also not a RE? I believe so, because we know that if $L$ is a RE then so is $L^r$. And so we end up with $(L')^rL^r$, and we can use the contrapositive of the pumping lemma to prove this language is not regular. Any insight would be helpful!










    share|cite|improve this question







    New contributor



    mathmathmath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$














      2












      2








      2





      $begingroup$


      I want to come up with a language that satisfies the pumping lemma while not being a regular expression.



      I thought of $0^i1^j: i > j > 0 $. The pumping seems to work just fine, and this is not a regular expression, as it would not be possible to create an automata that recognizes it. This last part is where I'm having a bit of trouble, as I think I have the right idea, but can't formalize it. This language is just a regular expression $big(0^i : i in mathbbNbig)$ concatenated with something that is not a regular expression $big(0^i1^i: i in mathbbNbig)$.



      Is my thought process correct? Can this be generalized? If $L$ is a RE and $L'$ is not, then $LL'$ is also not a RE? I believe so, because we know that if $L$ is a RE then so is $L^r$. And so we end up with $(L')^rL^r$, and we can use the contrapositive of the pumping lemma to prove this language is not regular. Any insight would be helpful!










      share|cite|improve this question







      New contributor



      mathmathmath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$




      I want to come up with a language that satisfies the pumping lemma while not being a regular expression.



      I thought of $0^i1^j: i > j > 0 $. The pumping seems to work just fine, and this is not a regular expression, as it would not be possible to create an automata that recognizes it. This last part is where I'm having a bit of trouble, as I think I have the right idea, but can't formalize it. This language is just a regular expression $big(0^i : i in mathbbNbig)$ concatenated with something that is not a regular expression $big(0^i1^i: i in mathbbNbig)$.



      Is my thought process correct? Can this be generalized? If $L$ is a RE and $L'$ is not, then $LL'$ is also not a RE? I believe so, because we know that if $L$ is a RE then so is $L^r$. And so we end up with $(L')^rL^r$, and we can use the contrapositive of the pumping lemma to prove this language is not regular. Any insight would be helpful!







      regular-languages regular-expressions pumping-lemma






      share|cite|improve this question







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      mathmathmath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.










      share|cite|improve this question







      New contributor



      mathmathmath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.








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      asked 8 hours ago









      mathmathmathmathmathmath

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          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Yes, $L_>=0^i1^j: i > j > 0 $ is a non-regular language.



          However, it does not satisfy the pumping lemma for regular languages. Assume for the sake of contradiction that $p$ is a pumping length for $L_>$. Consider $w=0^p+11^p$. Let $w=xyz$ with $|y|>1$ and $|xy|<p$. Then $y$ contains 0 only. There is no more 0's than 1's in $xy^0z=xz$. That contradicts that $p$ is a pumping length.



          The most we can say is that all words in $L_>$ can be pumped up.




          If $L$ is a RE and $L'$ is not, then $LL'$ is also not a RE?




          No. A simple counterexample is $L=Sigma^*$, the language of all words while $L'$ is any non-regular language that contains the empty word such as $a^n^2mid nin Bbb N_0$. $LL'$ is still the language of all words, which is regular.



          Exercise 1. Find two languages $A$ and $B$ such that




          • $A$ is regular


          • $B$ is not regular

          • $Asubsetneq B$


          • $AB$ is regular.

          Exercise 2. Find a non-regular language that satisfies the pumping lemma for regular languages.








          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Thank you for your answer, apreciate it a lot! In regards to Exercise 1, if you choose A = $varnothing$, then I believe it works :)
            $endgroup$
            – mathmathmath
            7 hours ago











          • $begingroup$
            @mathmathmath nice answer. To make it more difficult, I should have added $A$ is not empty.
            $endgroup$
            – Apass.Jack
            6 hours ago













          Your Answer








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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Yes, $L_>=0^i1^j: i > j > 0 $ is a non-regular language.



          However, it does not satisfy the pumping lemma for regular languages. Assume for the sake of contradiction that $p$ is a pumping length for $L_>$. Consider $w=0^p+11^p$. Let $w=xyz$ with $|y|>1$ and $|xy|<p$. Then $y$ contains 0 only. There is no more 0's than 1's in $xy^0z=xz$. That contradicts that $p$ is a pumping length.



          The most we can say is that all words in $L_>$ can be pumped up.




          If $L$ is a RE and $L'$ is not, then $LL'$ is also not a RE?




          No. A simple counterexample is $L=Sigma^*$, the language of all words while $L'$ is any non-regular language that contains the empty word such as $a^n^2mid nin Bbb N_0$. $LL'$ is still the language of all words, which is regular.



          Exercise 1. Find two languages $A$ and $B$ such that




          • $A$ is regular


          • $B$ is not regular

          • $Asubsetneq B$


          • $AB$ is regular.

          Exercise 2. Find a non-regular language that satisfies the pumping lemma for regular languages.








          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Thank you for your answer, apreciate it a lot! In regards to Exercise 1, if you choose A = $varnothing$, then I believe it works :)
            $endgroup$
            – mathmathmath
            7 hours ago











          • $begingroup$
            @mathmathmath nice answer. To make it more difficult, I should have added $A$ is not empty.
            $endgroup$
            – Apass.Jack
            6 hours ago















          2












          $begingroup$

          Yes, $L_>=0^i1^j: i > j > 0 $ is a non-regular language.



          However, it does not satisfy the pumping lemma for regular languages. Assume for the sake of contradiction that $p$ is a pumping length for $L_>$. Consider $w=0^p+11^p$. Let $w=xyz$ with $|y|>1$ and $|xy|<p$. Then $y$ contains 0 only. There is no more 0's than 1's in $xy^0z=xz$. That contradicts that $p$ is a pumping length.



          The most we can say is that all words in $L_>$ can be pumped up.




          If $L$ is a RE and $L'$ is not, then $LL'$ is also not a RE?




          No. A simple counterexample is $L=Sigma^*$, the language of all words while $L'$ is any non-regular language that contains the empty word such as $a^n^2mid nin Bbb N_0$. $LL'$ is still the language of all words, which is regular.



          Exercise 1. Find two languages $A$ and $B$ such that




          • $A$ is regular


          • $B$ is not regular

          • $Asubsetneq B$


          • $AB$ is regular.

          Exercise 2. Find a non-regular language that satisfies the pumping lemma for regular languages.








          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Thank you for your answer, apreciate it a lot! In regards to Exercise 1, if you choose A = $varnothing$, then I believe it works :)
            $endgroup$
            – mathmathmath
            7 hours ago











          • $begingroup$
            @mathmathmath nice answer. To make it more difficult, I should have added $A$ is not empty.
            $endgroup$
            – Apass.Jack
            6 hours ago













          2












          2








          2





          $begingroup$

          Yes, $L_>=0^i1^j: i > j > 0 $ is a non-regular language.



          However, it does not satisfy the pumping lemma for regular languages. Assume for the sake of contradiction that $p$ is a pumping length for $L_>$. Consider $w=0^p+11^p$. Let $w=xyz$ with $|y|>1$ and $|xy|<p$. Then $y$ contains 0 only. There is no more 0's than 1's in $xy^0z=xz$. That contradicts that $p$ is a pumping length.



          The most we can say is that all words in $L_>$ can be pumped up.




          If $L$ is a RE and $L'$ is not, then $LL'$ is also not a RE?




          No. A simple counterexample is $L=Sigma^*$, the language of all words while $L'$ is any non-regular language that contains the empty word such as $a^n^2mid nin Bbb N_0$. $LL'$ is still the language of all words, which is regular.



          Exercise 1. Find two languages $A$ and $B$ such that




          • $A$ is regular


          • $B$ is not regular

          • $Asubsetneq B$


          • $AB$ is regular.

          Exercise 2. Find a non-regular language that satisfies the pumping lemma for regular languages.








          share|cite|improve this answer











          $endgroup$



          Yes, $L_>=0^i1^j: i > j > 0 $ is a non-regular language.



          However, it does not satisfy the pumping lemma for regular languages. Assume for the sake of contradiction that $p$ is a pumping length for $L_>$. Consider $w=0^p+11^p$. Let $w=xyz$ with $|y|>1$ and $|xy|<p$. Then $y$ contains 0 only. There is no more 0's than 1's in $xy^0z=xz$. That contradicts that $p$ is a pumping length.



          The most we can say is that all words in $L_>$ can be pumped up.




          If $L$ is a RE and $L'$ is not, then $LL'$ is also not a RE?




          No. A simple counterexample is $L=Sigma^*$, the language of all words while $L'$ is any non-regular language that contains the empty word such as $a^n^2mid nin Bbb N_0$. $LL'$ is still the language of all words, which is regular.



          Exercise 1. Find two languages $A$ and $B$ such that




          • $A$ is regular


          • $B$ is not regular

          • $Asubsetneq B$


          • $AB$ is regular.

          Exercise 2. Find a non-regular language that satisfies the pumping lemma for regular languages.









          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 7 hours ago

























          answered 7 hours ago









          Apass.JackApass.Jack

          17.8k1 gold badge12 silver badges47 bronze badges




          17.8k1 gold badge12 silver badges47 bronze badges











          • $begingroup$
            Thank you for your answer, apreciate it a lot! In regards to Exercise 1, if you choose A = $varnothing$, then I believe it works :)
            $endgroup$
            – mathmathmath
            7 hours ago











          • $begingroup$
            @mathmathmath nice answer. To make it more difficult, I should have added $A$ is not empty.
            $endgroup$
            – Apass.Jack
            6 hours ago
















          • $begingroup$
            Thank you for your answer, apreciate it a lot! In regards to Exercise 1, if you choose A = $varnothing$, then I believe it works :)
            $endgroup$
            – mathmathmath
            7 hours ago











          • $begingroup$
            @mathmathmath nice answer. To make it more difficult, I should have added $A$ is not empty.
            $endgroup$
            – Apass.Jack
            6 hours ago















          $begingroup$
          Thank you for your answer, apreciate it a lot! In regards to Exercise 1, if you choose A = $varnothing$, then I believe it works :)
          $endgroup$
          – mathmathmath
          7 hours ago





          $begingroup$
          Thank you for your answer, apreciate it a lot! In regards to Exercise 1, if you choose A = $varnothing$, then I believe it works :)
          $endgroup$
          – mathmathmath
          7 hours ago













          $begingroup$
          @mathmathmath nice answer. To make it more difficult, I should have added $A$ is not empty.
          $endgroup$
          – Apass.Jack
          6 hours ago




          $begingroup$
          @mathmathmath nice answer. To make it more difficult, I should have added $A$ is not empty.
          $endgroup$
          – Apass.Jack
          6 hours ago










          mathmathmath is a new contributor. Be nice, and check out our Code of Conduct.









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