Mixing
How to disprove that the union of all non-context-free languages over $a$ is also non-context-free language?
$begingroup$
This is the proof I came across:
$L={a^{x^j}|jge 0, 2le xin mathbb N}$ is a non-context-free language. Suppose otherwise, then let $n$ be the constant promised in the pumping lemma. Let's choose $z=a^{x^n}, 1<x, zin L$ therefore exists a decomposition of $z$ into the following form: $z=uvwxy, u=a^{k_1},v=a^{k_2},w=a^{k_3},x=a^{k_4},y=a^{x^n-k_1-k_2-k_3-k_4}$
For $i=2: a^{x^n+k_2+k_4}quad$ *
$$
x^n< x^n+k_2+k_4le x^n+n<x^n+x^nle x^{n+1}quad **
$$
which means that the language is not context-free.
$L$ is contained in the union of all non-context-free languages and the union can be represented by the regular expression $a^+$ therefore the union is a context-free language (because it can be represented by reg. expression).
1) First I don't understand why we needed to bother with proving that $L$ is not context-free. Why couldn't we say from the very beginning that the union of all non-context-free languages is $a^+$ and therefore it's context-free?
2) Why for $i=2: a^{x^n+k_2+k_4}$? Is it because the length of $a^{x^n}$ without $k_2, k_4$ is $a^{x^n-k_2-k_4}$ and because $i=2$ then $a^{x^n-k_2-k_4+(2k_2+2k_4)}=a^{x^n+k_2+k_4}$?
3) In point ** we found an accurate upper bound for $x^n$ which is $x^{n+1}$. But it doesn't necessarily mean than the power of $x$ will be $n+1$ it's just an upper bound. Why is it enough for the proof?
proof-explanation formal-languages context-free-grammar
asked Jan 24 at 16:48
YosYos
1,1631823
$endgroup$
add a comment |
$begingroup$
This is the proof I came across:
$L={a^{x^j}|jge 0, 2le xin mathbb N}$ is a non-context-free language. Suppose otherwise, then let $n$ be the constant promised in the pumping lemma. Let's choose $z=a^{x^n}, 1<x, zin L$ therefore exists a decomposition of $z$ into the following form: $z=uvwxy, u=a^{k_1},v=a^{k_2},w=a^{k_3},x=a^{k_4},y=a^{x^n-k_1-k_2-k_3-k_4}$
For $i=2: a^{x^n+k_2+k_4}quad$ *
$$
x^n< x^n+k_2+k_4le x^n+n<x^n+x^nle x^{n+1}quad **
$$
which means that the language is not context-free.
$L$ is contained in the union of all non-context-free languages and the union can be represented by the regular expression $a^+$ therefore the union is a context-free language (because it can be represented by reg. expression).
1) First I don't understand why we needed to bother with proving that $L$ is not context-free. Why couldn't we say from the very beginning that the union of all non-context-free languages is $a^+$ and therefore it's context-free?
2) Why for $i=2: a^{x^n+k_2+k_4}$? Is it because the length of $a^{x^n}$ without $k_2, k_4$ is $a^{x^n-k_2-k_4}$ and because $i=2$ then $a^{x^n-k_2-k_4+(2k_2+2k_4)}=a^{x^n+k_2+k_4}$?
3) In point ** we found an accurate upper bound for $x^n$ which is $x^{n+1}$. But it doesn't necessarily mean than the power of $x$ will be $n+1$ it's just an upper bound. Why is it enough for the proof?
proof-explanation formal-languages context-free-grammar
asked Jan 24 at 16:48
YosYos
1,1631823
$endgroup$
add a comment |
$begingroup$
This is the proof I came across:
$L={a^{x^j}|jge 0, 2le xin mathbb N}$ is a non-context-free language. Suppose otherwise, then let $n$ be the constant promised in the pumping lemma. Let's choose $z=a^{x^n}, 1<x, zin L$ therefore exists a decomposition of $z$ into the following form: $z=uvwxy, u=a^{k_1},v=a^{k_2},w=a^{k_3},x=a^{k_4},y=a^{x^n-k_1-k_2-k_3-k_4}$
For $i=2: a^{x^n+k_2+k_4}quad$ *
$$
x^n< x^n+k_2+k_4le x^n+n<x^n+x^nle x^{n+1}quad **
$$
which means that the language is not context-free.
$L$ is contained in the union of all non-context-free languages and the union can be represented by the regular expression $a^+$ therefore the union is a context-free language (because it can be represented by reg. expression).
1) First I don't understand why we needed to bother with proving that $L$ is not context-free. Why couldn't we say from the very beginning that the union of all non-context-free languages is $a^+$ and therefore it's context-free?
2) Why for $i=2: a^{x^n+k_2+k_4}$? Is it because the length of $a^{x^n}$ without $k_2, k_4$ is $a^{x^n-k_2-k_4}$ and because $i=2$ then $a^{x^n-k_2-k_4+(2k_2+2k_4)}=a^{x^n+k_2+k_4}$?
3) In point ** we found an accurate upper bound for $x^n$ which is $x^{n+1}$. But it doesn't necessarily mean than the power of $x$ will be $n+1$ it's just an upper bound. Why is it enough for the proof?
proof-explanation formal-languages context-free-grammar
asked Jan 24 at 16:48
YosYos
1,1631823
$endgroup$
This is the proof I came across:
$L={a^{x^j}|jge 0, 2le xin mathbb N}$ is a non-context-free language. Suppose otherwise, then let $n$ be the constant promised in the pumping lemma. Let's choose $z=a^{x^n}, 1<x, zin L$ therefore exists a decomposition of $z$ into the following form: $z=uvwxy, u=a^{k_1},v=a^{k_2},w=a^{k_3},x=a^{k_4},y=a^{x^n-k_1-k_2-k_3-k_4}$
For $i=2: a^{x^n+k_2+k_4}quad$ *
$$
x^n< x^n+k_2+k_4le x^n+n<x^n+x^nle x^{n+1}quad **
$$
which means that the language is not context-free.
$L$ is contained in the union of all non-context-free languages and the union can be represented by the regular expression $a^+$ therefore the union is a context-free language (because it can be represented by reg. expression).
1) First I don't understand why we needed to bother with proving that $L$ is not context-free. Why couldn't we say from the very beginning that the union of all non-context-free languages is $a^+$ and therefore it's context-free?
2) Why for $i=2: a^{x^n+k_2+k_4}$? Is it because the length of $a^{x^n}$ without $k_2, k_4$ is $a^{x^n-k_2-k_4}$ and because $i=2$ then $a^{x^n-k_2-k_4+(2k_2+2k_4)}=a^{x^n+k_2+k_4}$?
3) In point ** we found an accurate upper bound for $x^n$ which is $x^{n+1}$. But it doesn't necessarily mean than the power of $x$ will be $n+1$ it's just an upper bound. Why is it enough for the proof?
proof-explanation formal-languages context-free-grammar
proof-explanation formal-languages context-free-grammar
asked Jan 24 at 16:48
YosYos
1,1631823
asked Jan 24 at 16:48
YosYos
1,1631823
edited Jan 28 at 16:22
Yos
asked Jan 24 at 16:48
YosYos
1,1631823
asked Jan 24 at 16:48
YosYos
1,1631823
asked Jan 24 at 16:48
YosYos
1,1631823
1,1631823
add a comment |
add a comment |
1 Answer
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$begingroup$
I think you made a mistake in the definition of L.
The x should be fixed once and for all. Otherwise $L=a*$ and all your question just don't make sens.
Thus I assumed that you wanted to show that $L_x={ a^{x^j}|jgeq 0} $ is non-context-free for $xinmathbb{N}$.
1) How do you know that " the union of all non-context-free languages is $a^+$"? I didn't know before knowing that the languages $L_x$ are non-context-free ...
2) yes
3) I think your confusion comes the confusion on the language definition. With the definition $L_x$ it should be clearer. If it is not please tell me.
answered Jan 28 at 15:49
wecewece
2,3721923
$endgroup$
$begingroup$
Yes $x$ should be fixed but it can be any natural number. In my first subquestion I meant why $a^{x^j}$ accurately describes the language but $a^+$ does not.
$endgroup$
– Yos
Jan 28 at 16:19
$begingroup$
I did understand the 3) point though now. Reading the inequality from right to left suddenly made more sense.
$endgroup$
– Yos
Jan 28 at 16:26
$begingroup$
if you're interested I have an interesting question here: cs.stackexchange.com/questions/103502/…
$endgroup$
– Yos
Jan 28 at 16:27
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I think you made a mistake in the definition of L.
The x should be fixed once and for all. Otherwise $L=a*$ and all your question just don't make sens.
Thus I assumed that you wanted to show that $L_x={ a^{x^j}|jgeq 0} $ is non-context-free for $xinmathbb{N}$.
1) How do you know that " the union of all non-context-free languages is $a^+$"? I didn't know before knowing that the languages $L_x$ are non-context-free ...
2) yes
3) I think your confusion comes the confusion on the language definition. With the definition $L_x$ it should be clearer. If it is not please tell me.
answered Jan 28 at 15:49
wecewece
2,3721923
$endgroup$
$begingroup$
Yes $x$ should be fixed but it can be any natural number. In my first subquestion I meant why $a^{x^j}$ accurately describes the language but $a^+$ does not.
$endgroup$
– Yos
Jan 28 at 16:19
$begingroup$
I did understand the 3) point though now. Reading the inequality from right to left suddenly made more sense.
$endgroup$
– Yos
Jan 28 at 16:26
$begingroup$
if you're interested I have an interesting question here: cs.stackexchange.com/questions/103502/…
$endgroup$
– Yos
Jan 28 at 16:27
add a comment |
$begingroup$
I think you made a mistake in the definition of L.
The x should be fixed once and for all. Otherwise $L=a*$ and all your question just don't make sens.
Thus I assumed that you wanted to show that $L_x={ a^{x^j}|jgeq 0} $ is non-context-free for $xinmathbb{N}$.
1) How do you know that " the union of all non-context-free languages is $a^+$"? I didn't know before knowing that the languages $L_x$ are non-context-free ...
2) yes
3) I think your confusion comes the confusion on the language definition. With the definition $L_x$ it should be clearer. If it is not please tell me.
answered Jan 28 at 15:49
wecewece
2,3721923
$endgroup$
$begingroup$
Yes $x$ should be fixed but it can be any natural number. In my first subquestion I meant why $a^{x^j}$ accurately describes the language but $a^+$ does not.
$endgroup$
– Yos
Jan 28 at 16:19
$begingroup$
I did understand the 3) point though now. Reading the inequality from right to left suddenly made more sense.
$endgroup$
– Yos
Jan 28 at 16:26
$begingroup$
if you're interested I have an interesting question here: cs.stackexchange.com/questions/103502/…
$endgroup$
– Yos
Jan 28 at 16:27
add a comment |
$begingroup$
I think you made a mistake in the definition of L.
The x should be fixed once and for all. Otherwise $L=a*$ and all your question just don't make sens.
Thus I assumed that you wanted to show that $L_x={ a^{x^j}|jgeq 0} $ is non-context-free for $xinmathbb{N}$.
1) How do you know that " the union of all non-context-free languages is $a^+$"? I didn't know before knowing that the languages $L_x$ are non-context-free ...
2) yes
3) I think your confusion comes the confusion on the language definition. With the definition $L_x$ it should be clearer. If it is not please tell me.
answered Jan 28 at 15:49
wecewece
2,3721923
$endgroup$
I think you made a mistake in the definition of L.
The x should be fixed once and for all. Otherwise $L=a*$ and all your question just don't make sens.
Thus I assumed that you wanted to show that $L_x={ a^{x^j}|jgeq 0} $ is non-context-free for $xinmathbb{N}$.
1) How do you know that " the union of all non-context-free languages is $a^+$"? I didn't know before knowing that the languages $L_x$ are non-context-free ...
2) yes
3) I think your confusion comes the confusion on the language definition. With the definition $L_x$ it should be clearer. If it is not please tell me.
answered Jan 28 at 15:49
wecewece
2,3721923
answered Jan 28 at 15:49
wecewece
2,3721923
answered Jan 28 at 15:49
wecewece
2,3721923
answered Jan 28 at 15:49
wecewece
2,3721923
2,3721923
$begingroup$
Yes $x$ should be fixed but it can be any natural number. In my first subquestion I meant why $a^{x^j}$ accurately describes the language but $a^+$ does not.
$endgroup$
– Yos
Jan 28 at 16:19
$begingroup$
I did understand the 3) point though now. Reading the inequality from right to left suddenly made more sense.
$endgroup$
– Yos
Jan 28 at 16:26
$begingroup$
if you're interested I have an interesting question here: cs.stackexchange.com/questions/103502/…
$endgroup$
– Yos
Jan 28 at 16:27
add a comment |
$begingroup$
Yes $x$ should be fixed but it can be any natural number. In my first subquestion I meant why $a^{x^j}$ accurately describes the language but $a^+$ does not.
$endgroup$
– Yos
Jan 28 at 16:19
$begingroup$
I did understand the 3) point though now. Reading the inequality from right to left suddenly made more sense.
$endgroup$
– Yos
Jan 28 at 16:26
$begingroup$
if you're interested I have an interesting question here: cs.stackexchange.com/questions/103502/…
$endgroup$
– Yos
Jan 28 at 16:27
$begingroup$
Yes $x$ should be fixed but it can be any natural number. In my first subquestion I meant why $a^{x^j}$ accurately describes the language but $a^+$ does not.
$endgroup$
– Yos
Jan 28 at 16:19
$begingroup$
Yes $x$ should be fixed but it can be any natural number. In my first subquestion I meant why $a^{x^j}$ accurately describes the language but $a^+$ does not.
$endgroup$
– Yos
Jan 28 at 16:19
$begingroup$
I did understand the 3) point though now. Reading the inequality from right to left suddenly made more sense.
$endgroup$
– Yos
Jan 28 at 16:26
$begingroup$
I did understand the 3) point though now. Reading the inequality from right to left suddenly made more sense.
$endgroup$
– Yos
Jan 28 at 16:26
$begingroup$
if you're interested I have an interesting question here: cs.stackexchange.com/questions/103502/…
$endgroup$
– Yos
Jan 28 at 16:27
$begingroup$
if you're interested I have an interesting question here: cs.stackexchange.com/questions/103502/…
$endgroup$
– Yos
Jan 28 at 16:27
add a comment |
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