How to prove that $L={a^kb^mc^{m-k}|mge kge0, m-kge k}$ is not context-free language?
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Prove that $L={a^kb^mc^{m-k}|mge kge0, m-kge k}$ is not context-free language.
We can suppose by contradiction that $L$ is context-free and choose $Z=a^kb^{2k}c^k$.
Using pumping lemma, $vwx$ can't have both $a$'s and $c$'s because $|vwx|le k$. There're $3$ cases:
1) $anotin vx, |vx|ge 1$ and $bin vx$. For $i=0:$ $$
Z_0=uv^0wx^0y=a^kb^{2k-l}c^{k-t}\lge 1\tge 0
$$ In this case $2k-lle 2kimplies Z_0notin L$.
2) $anotin vx, |vx|ge 1$ and $bnotin vx$ and $cin vx, |vx|ge 1$. For $i=0:$ $$
i=0: Z_0= uv^0wx^0y=a^kb^{2k-t}c^{k-l}\lge 1\tge 0
$$
In this case $k-l<kimplies Z_0notin L$
3) $ain vx, |vwx|le k, cnotin vx$. For $i=0:$
$$
Z_0=uv^0wx^0y=a^{k-l}2^{2k-t}c^k
$$
In this case $k-l<kimplies Z_0notin L$
I'm not sure that my proof good. If not what are my mistakes?
proof-writing context-free-grammar pumping-lemma
asked Dec 16 '18 at 18:52
YosYos
1,1631823
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add a comment |
$begingroup$
Prove that $L={a^kb^mc^{m-k}|mge kge0, m-kge k}$ is not context-free language.
We can suppose by contradiction that $L$ is context-free and choose $Z=a^kb^{2k}c^k$.
Using pumping lemma, $vwx$ can't have both $a$'s and $c$'s because $|vwx|le k$. There're $3$ cases:
1) $anotin vx, |vx|ge 1$ and $bin vx$. For $i=0:$ $$
Z_0=uv^0wx^0y=a^kb^{2k-l}c^{k-t}\lge 1\tge 0
$$ In this case $2k-lle 2kimplies Z_0notin L$.
2) $anotin vx, |vx|ge 1$ and $bnotin vx$ and $cin vx, |vx|ge 1$. For $i=0:$ $$
i=0: Z_0= uv^0wx^0y=a^kb^{2k-t}c^{k-l}\lge 1\tge 0
$$
In this case $k-l<kimplies Z_0notin L$
3) $ain vx, |vwx|le k, cnotin vx$. For $i=0:$
$$
Z_0=uv^0wx^0y=a^{k-l}2^{2k-t}c^k
$$
In this case $k-l<kimplies Z_0notin L$
I'm not sure that my proof good. If not what are my mistakes?
proof-writing context-free-grammar pumping-lemma
asked Dec 16 '18 at 18:52
YosYos
1,1631823
$endgroup$
add a comment |
$begingroup$
Prove that $L={a^kb^mc^{m-k}|mge kge0, m-kge k}$ is not context-free language.
We can suppose by contradiction that $L$ is context-free and choose $Z=a^kb^{2k}c^k$.
Using pumping lemma, $vwx$ can't have both $a$'s and $c$'s because $|vwx|le k$. There're $3$ cases:
1) $anotin vx, |vx|ge 1$ and $bin vx$. For $i=0:$ $$
Z_0=uv^0wx^0y=a^kb^{2k-l}c^{k-t}\lge 1\tge 0
$$ In this case $2k-lle 2kimplies Z_0notin L$.
2) $anotin vx, |vx|ge 1$ and $bnotin vx$ and $cin vx, |vx|ge 1$. For $i=0:$ $$
i=0: Z_0= uv^0wx^0y=a^kb^{2k-t}c^{k-l}\lge 1\tge 0
$$
In this case $k-l<kimplies Z_0notin L$
3) $ain vx, |vwx|le k, cnotin vx$. For $i=0:$
$$
Z_0=uv^0wx^0y=a^{k-l}2^{2k-t}c^k
$$
In this case $k-l<kimplies Z_0notin L$
I'm not sure that my proof good. If not what are my mistakes?
proof-writing context-free-grammar pumping-lemma
asked Dec 16 '18 at 18:52
YosYos
1,1631823
$endgroup$
Prove that $L={a^kb^mc^{m-k}|mge kge0, m-kge k}$ is not context-free language.
We can suppose by contradiction that $L$ is context-free and choose $Z=a^kb^{2k}c^k$.
Using pumping lemma, $vwx$ can't have both $a$'s and $c$'s because $|vwx|le k$. There're $3$ cases:
1) $anotin vx, |vx|ge 1$ and $bin vx$. For $i=0:$ $$
Z_0=uv^0wx^0y=a^kb^{2k-l}c^{k-t}\lge 1\tge 0
$$ In this case $2k-lle 2kimplies Z_0notin L$.
2) $anotin vx, |vx|ge 1$ and $bnotin vx$ and $cin vx, |vx|ge 1$. For $i=0:$ $$
i=0: Z_0= uv^0wx^0y=a^kb^{2k-t}c^{k-l}\lge 1\tge 0
$$
In this case $k-l<kimplies Z_0notin L$
3) $ain vx, |vwx|le k, cnotin vx$. For $i=0:$
$$
Z_0=uv^0wx^0y=a^{k-l}2^{2k-t}c^k
$$
In this case $k-l<kimplies Z_0notin L$
I'm not sure that my proof good. If not what are my mistakes?
proof-writing context-free-grammar pumping-lemma
proof-writing context-free-grammar pumping-lemma
asked Dec 16 '18 at 18:52
YosYos
1,1631823
asked Dec 16 '18 at 18:52
YosYos
1,1631823
edited Dec 21 '18 at 16:05
Yos
asked Dec 16 '18 at 18:52
YosYos
1,1631823
asked Dec 16 '18 at 18:52
YosYos
1,1631823
asked Dec 16 '18 at 18:52
YosYos
1,1631823
1,1631823
add a comment |
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