Prove that $L= $ $w $ ends with a palindrome of length greater than or equal to $4$ is nonregular using the pumping lemma.

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Hi, I tried this:



Assume to the contrary that $L$ is regular. Let $p$ be the pumping length given by the pumping lemma. Let $s$ be the string $a^pba^p$. Because $s$ is a member of $L$ and $s$ has length more than $p$, the pumping lemma guarantees that $s$ can be split into three pieces, $s=xyz$, satisfying the three conditions of the lemma:



1. for each $ige0,xy^izin L,$



2. $|y|>0, and$



3. $|xy| le p.$



$x=a^s, y=a^t,z=a^p-s-tba^p$



for $i=0,$ $xy^0z in L$



I don't understand how to solve it.



Thanks.




automata regular-language pumping-lemma




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    up vote
    0
    down vote

    favorite
    1












    The alphabet is $a, b$



    Hi, I tried this:



    Assume to the contrary that $L$ is regular. Let $p$ be the pumping length given by the pumping lemma. Let $s$ be the string $a^pba^p$. Because $s$ is a member of $L$ and $s$ has length more than $p$, the pumping lemma guarantees that $s$ can be split into three pieces, $s=xyz$, satisfying the three conditions of the lemma:



    1. for each $ige0,xy^izin L,$



    2. $|y|>0, and$



    3. $|xy| le p.$



    $x=a^s, y=a^t,z=a^p-s-tba^p$



    for $i=0,$ $xy^0z in L$



    I don't understand how to solve it.



    Thanks.




    automata regular-language pumping-lemma




    share|cite|improve this question





















      up vote
      0
      down vote

      favorite
      1









      up vote
      0
      down vote

      favorite
      1






      1





      The alphabet is $a, b$



      Hi, I tried this:



      Assume to the contrary that $L$ is regular. Let $p$ be the pumping length given by the pumping lemma. Let $s$ be the string $a^pba^p$. Because $s$ is a member of $L$ and $s$ has length more than $p$, the pumping lemma guarantees that $s$ can be split into three pieces, $s=xyz$, satisfying the three conditions of the lemma:



      1. for each $ige0,xy^izin L,$



      2. $|y|>0, and$



      3. $|xy| le p.$



      $x=a^s, y=a^t,z=a^p-s-tba^p$



      for $i=0,$ $xy^0z in L$



      I don't understand how to solve it.



      Thanks.




      automata regular-language pumping-lemma




      share|cite|improve this question











      The alphabet is $a, b$



      Hi, I tried this:



      Assume to the contrary that $L$ is regular. Let $p$ be the pumping length given by the pumping lemma. Let $s$ be the string $a^pba^p$. Because $s$ is a member of $L$ and $s$ has length more than $p$, the pumping lemma guarantees that $s$ can be split into three pieces, $s=xyz$, satisfying the three conditions of the lemma:



      1. for each $ige0,xy^izin L,$



      2. $|y|>0, and$



      3. $|xy| le p.$



      $x=a^s, y=a^t,z=a^p-s-tba^p$



      for $i=0,$ $xy^0z in L$



      I don't understand how to solve it.



      Thanks.





      automata regular-language pumping-lemma





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      share|cite|improve this question




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