Proof verification of the language of all palindromes as being context-free












0












$begingroup$


Consider that the language L of all palindromes over $Sigma = {0,1}^*$ is not context-free. The following is my attempt at a proof by contradiction.



I am new to proof writing and I am wondering if the proof is correct, and if it proceeds in a connected logical sequence. I think I have all the cases covered, but I am not too sure.



enter image description here






proof-verification proof-writing context-free-grammar pumping-lemma palindrome







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$endgroup$








  • 3




    $begingroup$
    Could you explain why $uwy = 0^k1^{2k}o^k$? If I set $w = 1, v = x = 1$, then neither $x$ nor $v$ equals $epsilon$, $|vwx| < p$ and $uv^iwx^iy in L$, but $uwy$ is not of the form you state. Apart from that, there's a very simple grammar for palindromes: $w := "" | "1" | "0" | "1" w "1" | "0" w "0"$ that looks pretty context free to me
    $endgroup$
    – Ronald
    Dec 6 '18 at 0:33










  • $begingroup$
    @Ronald could you elaborate? Are you saying that the language of palindromes is context free?
    $endgroup$
    – SeesSound
    Dec 6 '18 at 1:02


















0












$begingroup$


Consider that the language L of all palindromes over $Sigma = {0,1}^*$ is not context-free. The following is my attempt at a proof by contradiction.



I am new to proof writing and I am wondering if the proof is correct, and if it proceeds in a connected logical sequence. I think I have all the cases covered, but I am not too sure.



enter image description here






proof-verification proof-writing context-free-grammar pumping-lemma palindrome







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Could you explain why $uwy = 0^k1^{2k}o^k$? If I set $w = 1, v = x = 1$, then neither $x$ nor $v$ equals $epsilon$, $|vwx| < p$ and $uv^iwx^iy in L$, but $uwy$ is not of the form you state. Apart from that, there's a very simple grammar for palindromes: $w := "" | "1" | "0" | "1" w "1" | "0" w "0"$ that looks pretty context free to me
    $endgroup$
    – Ronald
    Dec 6 '18 at 0:33










  • $begingroup$
    @Ronald could you elaborate? Are you saying that the language of palindromes is context free?
    $endgroup$
    – SeesSound
    Dec 6 '18 at 1:02
















0












0








0





$begingroup$


Consider that the language L of all palindromes over $Sigma = {0,1}^*$ is not context-free. The following is my attempt at a proof by contradiction.



I am new to proof writing and I am wondering if the proof is correct, and if it proceeds in a connected logical sequence. I think I have all the cases covered, but I am not too sure.



enter image description here






proof-verification proof-writing context-free-grammar pumping-lemma palindrome







share|cite|improve this question









$endgroup$




Consider that the language L of all palindromes over $Sigma = {0,1}^*$ is not context-free. The following is my attempt at a proof by contradiction.



I am new to proof writing and I am wondering if the proof is correct, and if it proceeds in a connected logical sequence. I think I have all the cases covered, but I am not too sure.



enter image description here






proof-verification proof-writing context-free-grammar pumping-lemma palindrome




proof-verification proof-writing context-free-grammar pumping-lemma palindrome






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 23:57









SeesSoundSeesSound

859




859








  • 3




    $begingroup$
    Could you explain why $uwy = 0^k1^{2k}o^k$? If I set $w = 1, v = x = 1$, then neither $x$ nor $v$ equals $epsilon$, $|vwx| < p$ and $uv^iwx^iy in L$, but $uwy$ is not of the form you state. Apart from that, there's a very simple grammar for palindromes: $w := "" | "1" | "0" | "1" w "1" | "0" w "0"$ that looks pretty context free to me
    $endgroup$
    – Ronald
    Dec 6 '18 at 0:33










  • $begingroup$
    @Ronald could you elaborate? Are you saying that the language of palindromes is context free?
    $endgroup$
    – SeesSound
    Dec 6 '18 at 1:02
















  • 3




    $begingroup$
    Could you explain why $uwy = 0^k1^{2k}o^k$? If I set $w = 1, v = x = 1$, then neither $x$ nor $v$ equals $epsilon$, $|vwx| < p$ and $uv^iwx^iy in L$, but $uwy$ is not of the form you state. Apart from that, there's a very simple grammar for palindromes: $w := "" | "1" | "0" | "1" w "1" | "0" w "0"$ that looks pretty context free to me
    $endgroup$
    – Ronald
    Dec 6 '18 at 0:33










  • $begingroup$
    @Ronald could you elaborate? Are you saying that the language of palindromes is context free?
    $endgroup$
    – SeesSound
    Dec 6 '18 at 1:02










3




3




$begingroup$
Could you explain why $uwy = 0^k1^{2k}o^k$? If I set $w = 1, v = x = 1$, then neither $x$ nor $v$ equals $epsilon$, $|vwx| < p$ and $uv^iwx^iy in L$, but $uwy$ is not of the form you state. Apart from that, there's a very simple grammar for palindromes: $w := "" | "1" | "0" | "1" w "1" | "0" w "0"$ that looks pretty context free to me
$endgroup$
– Ronald
Dec 6 '18 at 0:33




$begingroup$
Could you explain why $uwy = 0^k1^{2k}o^k$? If I set $w = 1, v = x = 1$, then neither $x$ nor $v$ equals $epsilon$, $|vwx| < p$ and $uv^iwx^iy in L$, but $uwy$ is not of the form you state. Apart from that, there's a very simple grammar for palindromes: $w := "" | "1" | "0" | "1" w "1" | "0" w "0"$ that looks pretty context free to me
$endgroup$
– Ronald
Dec 6 '18 at 0:33












$begingroup$
@Ronald could you elaborate? Are you saying that the language of palindromes is context free?
$endgroup$
– SeesSound
Dec 6 '18 at 1:02






$begingroup$
@Ronald could you elaborate? Are you saying that the language of palindromes is context free?
$endgroup$
– SeesSound
Dec 6 '18 at 1:02












1 Answer
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$begingroup$

If a context free grammar exists that produces the language, then the language itself is context free.



A context free grammar is a 4-tuple $G = (V, Sigma, R, S)$ with the following properties:





  1. $V$ is a finite set; $v in V$ is called a non-terminal symbol


  2. $Sigma$ is a finite set of terminal symbols; $V cap Sigma =emptyset$


  3. $R$ is a finite relation $V rightarrow (V cup Sigma)^*$


  4. $S in V$ is the start symbol


With $V = {w}$, $Sigma = {0, 1}$, $S = w$ and the following map $R$:



$w rightarrow epsilon$
$w rightarrow 0$
$w rightarrow 1$
$w rightarrow 0w0$
$w rightarrow 1w1$



we've defined a context free grammar.



Note: if the empty string isn't considered to be a palindrome, the first production can be replaced by



$w rightarrow 00$
$w rightarrow 11$



Palindromes are, casually speaking, a special case of "well-formed parenthesis" (each opening parenthesis has a corresponding closing parenthesis) which is also known to be context free.



Context free languages can't "count" very well, but it isn't a problem to have an arbitrary number of matching pairs as this can be tested using a stack.
But the famous language $a^nb^nc^n$ isn't context free because there's no way to count to $n$.



Edit:

Strictly speaking I'll have to prove that the above grammar indeed produces the set of all palindromes.



First of all, if $w in L$, which means that $w$ is a palindrome, then also $0w0$ and $1w1$ are palindromes and therefore member of $L$.
Obviously $0, 1$ and $epsilon$ are palindromes.

In other words, the production rules only produce palindromes.



Now assume $p$ is a palindrome. Then $p = s_1s_2...s_nms_ns_{n-1}...s_1$, with $s_i in {0, 1}, i in {1, ..., n}$ and $m in {epsilon, 0, 1}$.



In order to produce this palindrome, we start with $w = m$ and then consecutively add the symbols $s_n, s_{n-1}, ..., s_1$ on either side of $w$, which is allowed by the last two production rules.



This proves that $p$ can be produced by the grammar above.

Together with the fact that $G$ produces palindromes only, this proves that $G$ is a valid grammar for $L$.






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    0












    $begingroup$

    If a context free grammar exists that produces the language, then the language itself is context free.



    A context free grammar is a 4-tuple $G = (V, Sigma, R, S)$ with the following properties:





    1. $V$ is a finite set; $v in V$ is called a non-terminal symbol


    2. $Sigma$ is a finite set of terminal symbols; $V cap Sigma =emptyset$


    3. $R$ is a finite relation $V rightarrow (V cup Sigma)^*$


    4. $S in V$ is the start symbol


    With $V = {w}$, $Sigma = {0, 1}$, $S = w$ and the following map $R$:



    $w rightarrow epsilon$
    $w rightarrow 0$
    $w rightarrow 1$
    $w rightarrow 0w0$
    $w rightarrow 1w1$



    we've defined a context free grammar.



    Note: if the empty string isn't considered to be a palindrome, the first production can be replaced by



    $w rightarrow 00$
    $w rightarrow 11$



    Palindromes are, casually speaking, a special case of "well-formed parenthesis" (each opening parenthesis has a corresponding closing parenthesis) which is also known to be context free.



    Context free languages can't "count" very well, but it isn't a problem to have an arbitrary number of matching pairs as this can be tested using a stack.
    But the famous language $a^nb^nc^n$ isn't context free because there's no way to count to $n$.



    Edit:

    Strictly speaking I'll have to prove that the above grammar indeed produces the set of all palindromes.



    First of all, if $w in L$, which means that $w$ is a palindrome, then also $0w0$ and $1w1$ are palindromes and therefore member of $L$.
    Obviously $0, 1$ and $epsilon$ are palindromes.

    In other words, the production rules only produce palindromes.



    Now assume $p$ is a palindrome. Then $p = s_1s_2...s_nms_ns_{n-1}...s_1$, with $s_i in {0, 1}, i in {1, ..., n}$ and $m in {epsilon, 0, 1}$.



    In order to produce this palindrome, we start with $w = m$ and then consecutively add the symbols $s_n, s_{n-1}, ..., s_1$ on either side of $w$, which is allowed by the last two production rules.



    This proves that $p$ can be produced by the grammar above.

    Together with the fact that $G$ produces palindromes only, this proves that $G$ is a valid grammar for $L$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      If a context free grammar exists that produces the language, then the language itself is context free.



      A context free grammar is a 4-tuple $G = (V, Sigma, R, S)$ with the following properties:





      1. $V$ is a finite set; $v in V$ is called a non-terminal symbol


      2. $Sigma$ is a finite set of terminal symbols; $V cap Sigma =emptyset$


      3. $R$ is a finite relation $V rightarrow (V cup Sigma)^*$


      4. $S in V$ is the start symbol


      With $V = {w}$, $Sigma = {0, 1}$, $S = w$ and the following map $R$:



      $w rightarrow epsilon$
      $w rightarrow 0$
      $w rightarrow 1$
      $w rightarrow 0w0$
      $w rightarrow 1w1$



      we've defined a context free grammar.



      Note: if the empty string isn't considered to be a palindrome, the first production can be replaced by



      $w rightarrow 00$
      $w rightarrow 11$



      Palindromes are, casually speaking, a special case of "well-formed parenthesis" (each opening parenthesis has a corresponding closing parenthesis) which is also known to be context free.



      Context free languages can't "count" very well, but it isn't a problem to have an arbitrary number of matching pairs as this can be tested using a stack.
      But the famous language $a^nb^nc^n$ isn't context free because there's no way to count to $n$.



      Edit:

      Strictly speaking I'll have to prove that the above grammar indeed produces the set of all palindromes.



      First of all, if $w in L$, which means that $w$ is a palindrome, then also $0w0$ and $1w1$ are palindromes and therefore member of $L$.
      Obviously $0, 1$ and $epsilon$ are palindromes.

      In other words, the production rules only produce palindromes.



      Now assume $p$ is a palindrome. Then $p = s_1s_2...s_nms_ns_{n-1}...s_1$, with $s_i in {0, 1}, i in {1, ..., n}$ and $m in {epsilon, 0, 1}$.



      In order to produce this palindrome, we start with $w = m$ and then consecutively add the symbols $s_n, s_{n-1}, ..., s_1$ on either side of $w$, which is allowed by the last two production rules.



      This proves that $p$ can be produced by the grammar above.

      Together with the fact that $G$ produces palindromes only, this proves that $G$ is a valid grammar for $L$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        If a context free grammar exists that produces the language, then the language itself is context free.



        A context free grammar is a 4-tuple $G = (V, Sigma, R, S)$ with the following properties:





        1. $V$ is a finite set; $v in V$ is called a non-terminal symbol


        2. $Sigma$ is a finite set of terminal symbols; $V cap Sigma =emptyset$


        3. $R$ is a finite relation $V rightarrow (V cup Sigma)^*$


        4. $S in V$ is the start symbol


        With $V = {w}$, $Sigma = {0, 1}$, $S = w$ and the following map $R$:



        $w rightarrow epsilon$
        $w rightarrow 0$
        $w rightarrow 1$
        $w rightarrow 0w0$
        $w rightarrow 1w1$



        we've defined a context free grammar.



        Note: if the empty string isn't considered to be a palindrome, the first production can be replaced by



        $w rightarrow 00$
        $w rightarrow 11$



        Palindromes are, casually speaking, a special case of "well-formed parenthesis" (each opening parenthesis has a corresponding closing parenthesis) which is also known to be context free.



        Context free languages can't "count" very well, but it isn't a problem to have an arbitrary number of matching pairs as this can be tested using a stack.
        But the famous language $a^nb^nc^n$ isn't context free because there's no way to count to $n$.



        Edit:

        Strictly speaking I'll have to prove that the above grammar indeed produces the set of all palindromes.



        First of all, if $w in L$, which means that $w$ is a palindrome, then also $0w0$ and $1w1$ are palindromes and therefore member of $L$.
        Obviously $0, 1$ and $epsilon$ are palindromes.

        In other words, the production rules only produce palindromes.



        Now assume $p$ is a palindrome. Then $p = s_1s_2...s_nms_ns_{n-1}...s_1$, with $s_i in {0, 1}, i in {1, ..., n}$ and $m in {epsilon, 0, 1}$.



        In order to produce this palindrome, we start with $w = m$ and then consecutively add the symbols $s_n, s_{n-1}, ..., s_1$ on either side of $w$, which is allowed by the last two production rules.



        This proves that $p$ can be produced by the grammar above.

        Together with the fact that $G$ produces palindromes only, this proves that $G$ is a valid grammar for $L$.






        share|cite|improve this answer











        $endgroup$



        If a context free grammar exists that produces the language, then the language itself is context free.



        A context free grammar is a 4-tuple $G = (V, Sigma, R, S)$ with the following properties:





        1. $V$ is a finite set; $v in V$ is called a non-terminal symbol


        2. $Sigma$ is a finite set of terminal symbols; $V cap Sigma =emptyset$


        3. $R$ is a finite relation $V rightarrow (V cup Sigma)^*$


        4. $S in V$ is the start symbol


        With $V = {w}$, $Sigma = {0, 1}$, $S = w$ and the following map $R$:



        $w rightarrow epsilon$
        $w rightarrow 0$
        $w rightarrow 1$
        $w rightarrow 0w0$
        $w rightarrow 1w1$



        we've defined a context free grammar.



        Note: if the empty string isn't considered to be a palindrome, the first production can be replaced by



        $w rightarrow 00$
        $w rightarrow 11$



        Palindromes are, casually speaking, a special case of "well-formed parenthesis" (each opening parenthesis has a corresponding closing parenthesis) which is also known to be context free.



        Context free languages can't "count" very well, but it isn't a problem to have an arbitrary number of matching pairs as this can be tested using a stack.
        But the famous language $a^nb^nc^n$ isn't context free because there's no way to count to $n$.



        Edit:

        Strictly speaking I'll have to prove that the above grammar indeed produces the set of all palindromes.



        First of all, if $w in L$, which means that $w$ is a palindrome, then also $0w0$ and $1w1$ are palindromes and therefore member of $L$.
        Obviously $0, 1$ and $epsilon$ are palindromes.

        In other words, the production rules only produce palindromes.



        Now assume $p$ is a palindrome. Then $p = s_1s_2...s_nms_ns_{n-1}...s_1$, with $s_i in {0, 1}, i in {1, ..., n}$ and $m in {epsilon, 0, 1}$.



        In order to produce this palindrome, we start with $w = m$ and then consecutively add the symbols $s_n, s_{n-1}, ..., s_1$ on either side of $w$, which is allowed by the last two production rules.



        This proves that $p$ can be produced by the grammar above.

        Together with the fact that $G$ produces palindromes only, this proves that $G$ is a valid grammar for $L$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 6 '18 at 14:02

























        answered Dec 6 '18 at 9:08









        RonaldRonald

        673510




        673510






























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