Is $L_4$ a CFL?












0












$begingroup$



Consider the following language: $$L_4 = {a^ib^jc^kd^l : i,j,k,l
ge0 wedge i=1 Rightarrow j=k=l}.$$
Prove or disprove: $L_4$ is a context-free language.




To me, it looks like $L_4$ can be accepted using a PDA, but I don't know how to construct it.
appreciate a hint here.






formal-languages automata context-free-grammar formal-grammar pumping-lemma







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    The sub-language for $i=1$ is not context.free
    $endgroup$
    – Wuestenfux
    Dec 22 '18 at 13:42










  • $begingroup$
    Can you give me a hint of how to disprove it then?
    $endgroup$
    – Yotam Raz
    Dec 22 '18 at 14:48






  • 1




    $begingroup$
    The pumping lemma for context-free languages would do it.
    $endgroup$
    – Henning Makholm
    Dec 22 '18 at 14:54










  • $begingroup$
    is $w = ab^kc^kd^k$ a good choice? for $k$ being the constant from the pumping lemma.
    $endgroup$
    – Yotam Raz
    Dec 22 '18 at 15:03








  • 1




    $begingroup$
    More precisely, what the pumping lemma shows directly is @Wuestenfux's claim. To get all the way to $L_4$, note that "the sub-language for $i=1$" is the intersection of $L_4$ and a regular language.
    $endgroup$
    – Henning Makholm
    Dec 22 '18 at 15:50
















0












$begingroup$



Consider the following language: $$L_4 = {a^ib^jc^kd^l : i,j,k,l
ge0 wedge i=1 Rightarrow j=k=l}.$$
Prove or disprove: $L_4$ is a context-free language.




To me, it looks like $L_4$ can be accepted using a PDA, but I don't know how to construct it.
appreciate a hint here.






formal-languages automata context-free-grammar formal-grammar pumping-lemma







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    The sub-language for $i=1$ is not context.free
    $endgroup$
    – Wuestenfux
    Dec 22 '18 at 13:42










  • $begingroup$
    Can you give me a hint of how to disprove it then?
    $endgroup$
    – Yotam Raz
    Dec 22 '18 at 14:48






  • 1




    $begingroup$
    The pumping lemma for context-free languages would do it.
    $endgroup$
    – Henning Makholm
    Dec 22 '18 at 14:54










  • $begingroup$
    is $w = ab^kc^kd^k$ a good choice? for $k$ being the constant from the pumping lemma.
    $endgroup$
    – Yotam Raz
    Dec 22 '18 at 15:03








  • 1




    $begingroup$
    More precisely, what the pumping lemma shows directly is @Wuestenfux's claim. To get all the way to $L_4$, note that "the sub-language for $i=1$" is the intersection of $L_4$ and a regular language.
    $endgroup$
    – Henning Makholm
    Dec 22 '18 at 15:50














0












0








0





$begingroup$



Consider the following language: $$L_4 = {a^ib^jc^kd^l : i,j,k,l
ge0 wedge i=1 Rightarrow j=k=l}.$$
Prove or disprove: $L_4$ is a context-free language.




To me, it looks like $L_4$ can be accepted using a PDA, but I don't know how to construct it.
appreciate a hint here.






formal-languages automata context-free-grammar formal-grammar pumping-lemma







share|cite|improve this question











$endgroup$





Consider the following language: $$L_4 = {a^ib^jc^kd^l : i,j,k,l
ge0 wedge i=1 Rightarrow j=k=l}.$$
Prove or disprove: $L_4$ is a context-free language.




To me, it looks like $L_4$ can be accepted using a PDA, but I don't know how to construct it.
appreciate a hint here.






formal-languages automata context-free-grammar formal-grammar pumping-lemma




formal-languages automata context-free-grammar formal-grammar pumping-lemma






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 21:21









Yuval Filmus

48.7k471145




48.7k471145










asked Dec 22 '18 at 11:53









Yotam RazYotam Raz

235




235








  • 3




    $begingroup$
    The sub-language for $i=1$ is not context.free
    $endgroup$
    – Wuestenfux
    Dec 22 '18 at 13:42










  • $begingroup$
    Can you give me a hint of how to disprove it then?
    $endgroup$
    – Yotam Raz
    Dec 22 '18 at 14:48






  • 1




    $begingroup$
    The pumping lemma for context-free languages would do it.
    $endgroup$
    – Henning Makholm
    Dec 22 '18 at 14:54










  • $begingroup$
    is $w = ab^kc^kd^k$ a good choice? for $k$ being the constant from the pumping lemma.
    $endgroup$
    – Yotam Raz
    Dec 22 '18 at 15:03








  • 1




    $begingroup$
    More precisely, what the pumping lemma shows directly is @Wuestenfux's claim. To get all the way to $L_4$, note that "the sub-language for $i=1$" is the intersection of $L_4$ and a regular language.
    $endgroup$
    – Henning Makholm
    Dec 22 '18 at 15:50














  • 3




    $begingroup$
    The sub-language for $i=1$ is not context.free
    $endgroup$
    – Wuestenfux
    Dec 22 '18 at 13:42










  • $begingroup$
    Can you give me a hint of how to disprove it then?
    $endgroup$
    – Yotam Raz
    Dec 22 '18 at 14:48






  • 1




    $begingroup$
    The pumping lemma for context-free languages would do it.
    $endgroup$
    – Henning Makholm
    Dec 22 '18 at 14:54










  • $begingroup$
    is $w = ab^kc^kd^k$ a good choice? for $k$ being the constant from the pumping lemma.
    $endgroup$
    – Yotam Raz
    Dec 22 '18 at 15:03








  • 1




    $begingroup$
    More precisely, what the pumping lemma shows directly is @Wuestenfux's claim. To get all the way to $L_4$, note that "the sub-language for $i=1$" is the intersection of $L_4$ and a regular language.
    $endgroup$
    – Henning Makholm
    Dec 22 '18 at 15:50








3




3




$begingroup$
The sub-language for $i=1$ is not context.free
$endgroup$
– Wuestenfux
Dec 22 '18 at 13:42




$begingroup$
The sub-language for $i=1$ is not context.free
$endgroup$
– Wuestenfux
Dec 22 '18 at 13:42












$begingroup$
Can you give me a hint of how to disprove it then?
$endgroup$
– Yotam Raz
Dec 22 '18 at 14:48




$begingroup$
Can you give me a hint of how to disprove it then?
$endgroup$
– Yotam Raz
Dec 22 '18 at 14:48




1




1




$begingroup$
The pumping lemma for context-free languages would do it.
$endgroup$
– Henning Makholm
Dec 22 '18 at 14:54




$begingroup$
The pumping lemma for context-free languages would do it.
$endgroup$
– Henning Makholm
Dec 22 '18 at 14:54












$begingroup$
is $w = ab^kc^kd^k$ a good choice? for $k$ being the constant from the pumping lemma.
$endgroup$
– Yotam Raz
Dec 22 '18 at 15:03






$begingroup$
is $w = ab^kc^kd^k$ a good choice? for $k$ being the constant from the pumping lemma.
$endgroup$
– Yotam Raz
Dec 22 '18 at 15:03






1




1




$begingroup$
More precisely, what the pumping lemma shows directly is @Wuestenfux's claim. To get all the way to $L_4$, note that "the sub-language for $i=1$" is the intersection of $L_4$ and a regular language.
$endgroup$
– Henning Makholm
Dec 22 '18 at 15:50




$begingroup$
More precisely, what the pumping lemma shows directly is @Wuestenfux's claim. To get all the way to $L_4$, note that "the sub-language for $i=1$" is the intersection of $L_4$ and a regular language.
$endgroup$
– Henning Makholm
Dec 22 '18 at 15:50










1 Answer
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$begingroup$

Since the context-free languages are closed under intersection with a regular language and under homomorphism, if $L_4$ were context-free then so would the following language be, for the homomorphism given by $h(a) = epsilon$ and $h(sigma) = sigma$ for $sigma neq a$:
$$
h(L_4 cap a(b+c+d)^*) = { b^nc^nd^n : n geq 0 }.
$$
However, the latter language is well-known not to be context-free.






share|cite|improve this answer











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    $begingroup$

    Since the context-free languages are closed under intersection with a regular language and under homomorphism, if $L_4$ were context-free then so would the following language be, for the homomorphism given by $h(a) = epsilon$ and $h(sigma) = sigma$ for $sigma neq a$:
    $$
    h(L_4 cap a(b+c+d)^*) = { b^nc^nd^n : n geq 0 }.
    $$
    However, the latter language is well-known not to be context-free.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Since the context-free languages are closed under intersection with a regular language and under homomorphism, if $L_4$ were context-free then so would the following language be, for the homomorphism given by $h(a) = epsilon$ and $h(sigma) = sigma$ for $sigma neq a$:
      $$
      h(L_4 cap a(b+c+d)^*) = { b^nc^nd^n : n geq 0 }.
      $$
      However, the latter language is well-known not to be context-free.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Since the context-free languages are closed under intersection with a regular language and under homomorphism, if $L_4$ were context-free then so would the following language be, for the homomorphism given by $h(a) = epsilon$ and $h(sigma) = sigma$ for $sigma neq a$:
        $$
        h(L_4 cap a(b+c+d)^*) = { b^nc^nd^n : n geq 0 }.
        $$
        However, the latter language is well-known not to be context-free.






        share|cite|improve this answer











        $endgroup$



        Since the context-free languages are closed under intersection with a regular language and under homomorphism, if $L_4$ were context-free then so would the following language be, for the homomorphism given by $h(a) = epsilon$ and $h(sigma) = sigma$ for $sigma neq a$:
        $$
        h(L_4 cap a(b+c+d)^*) = { b^nc^nd^n : n geq 0 }.
        $$
        However, the latter language is well-known not to be context-free.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 19 at 16:43

























        answered Jan 16 at 21:23









        Yuval FilmusYuval Filmus

        48.7k471145




        48.7k471145






























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