Mixing
How to prove that concatenation of words from regular languages is a regular language using left quotient?
$begingroup$
Let
$$
L={xin Sigma^*big|x=uvw, \u,v,w in Sigma^*,\uin L_1,\vin L_1,\uwin L_2}
$$
where $L_1, L_2$ are regular languages over $sum^*$. Prove that $L$ is also regular.
I'd like to prove this using closure properties. We can define $L_q=L_2 backslash L_1={w}$ (get left quotient). But in order to concatenate $uvw$ I need to extract somehow the language of all $u$'s and all $v$'s from $L_1$. I'm not sure though.
formal-languages regular-language
edited Jan 28 at 12:37
Peter Leupold
63526
asked Jan 25 at 7:32
YosYos
1,1631823
$endgroup$
add a comment |
$begingroup$
Let
$$
L={xin Sigma^*big|x=uvw, \u,v,w in Sigma^*,\uin L_1,\vin L_1,\uwin L_2}
$$
where $L_1, L_2$ are regular languages over $sum^*$. Prove that $L$ is also regular.
I'd like to prove this using closure properties. We can define $L_q=L_2 backslash L_1={w}$ (get left quotient). But in order to concatenate $uvw$ I need to extract somehow the language of all $u$'s and all $v$'s from $L_1$. I'm not sure though.
formal-languages regular-language
edited Jan 28 at 12:37
Peter Leupold
63526
asked Jan 25 at 7:32
YosYos
1,1631823
$endgroup$
1
$begingroup$
The $u$ are from the language right quotient of ($L_2$ by $L_1$) intersected with $L_1$; the $v$ are from $L_1$. But if you concatenate the three components, you cannot guarantee that the $u$ and the $w$ are from the same $uw$. It might be hard to prove this via closure properties. On the other hand, it is quite straight-forward via automata.
$endgroup$
– Peter Leupold
Jan 28 at 12:44
add a comment |
$begingroup$
Let
$$
L={xin Sigma^*big|x=uvw, \u,v,w in Sigma^*,\uin L_1,\vin L_1,\uwin L_2}
$$
where $L_1, L_2$ are regular languages over $sum^*$. Prove that $L$ is also regular.
I'd like to prove this using closure properties. We can define $L_q=L_2 backslash L_1={w}$ (get left quotient). But in order to concatenate $uvw$ I need to extract somehow the language of all $u$'s and all $v$'s from $L_1$. I'm not sure though.
formal-languages regular-language
edited Jan 28 at 12:37
Peter Leupold
63526
asked Jan 25 at 7:32
YosYos
1,1631823
$endgroup$
Let
$$
L={xin Sigma^*big|x=uvw, \u,v,w in Sigma^*,\uin L_1,\vin L_1,\uwin L_2}
$$
where $L_1, L_2$ are regular languages over $sum^*$. Prove that $L$ is also regular.
I'd like to prove this using closure properties. We can define $L_q=L_2 backslash L_1={w}$ (get left quotient). But in order to concatenate $uvw$ I need to extract somehow the language of all $u$'s and all $v$'s from $L_1$. I'm not sure though.
formal-languages regular-language
formal-languages regular-language
edited Jan 28 at 12:37
Peter Leupold
63526
asked Jan 25 at 7:32
YosYos
1,1631823
edited Jan 28 at 12:37
Peter Leupold
63526
asked Jan 25 at 7:32
YosYos
1,1631823
edited Jan 28 at 12:37
Peter Leupold
63526
edited Jan 28 at 12:37
Peter Leupold
63526
edited Jan 28 at 12:37
Peter Leupold
63526
63526
asked Jan 25 at 7:32
YosYos
1,1631823
asked Jan 25 at 7:32
YosYos
1,1631823
asked Jan 25 at 7:32
YosYos
1,1631823
1,1631823
1
$begingroup$
The $u$ are from the language right quotient of ($L_2$ by $L_1$) intersected with $L_1$; the $v$ are from $L_1$. But if you concatenate the three components, you cannot guarantee that the $u$ and the $w$ are from the same $uw$. It might be hard to prove this via closure properties. On the other hand, it is quite straight-forward via automata.
$endgroup$
– Peter Leupold
Jan 28 at 12:44
add a comment |
1
$begingroup$
The $u$ are from the language right quotient of ($L_2$ by $L_1$) intersected with $L_1$; the $v$ are from $L_1$. But if you concatenate the three components, you cannot guarantee that the $u$ and the $w$ are from the same $uw$. It might be hard to prove this via closure properties. On the other hand, it is quite straight-forward via automata.
$endgroup$
– Peter Leupold
Jan 28 at 12:44
1
1
$begingroup$
The $u$ are from the language right quotient of ($L_2$ by $L_1$) intersected with $L_1$; the $v$ are from $L_1$. But if you concatenate the three components, you cannot guarantee that the $u$ and the $w$ are from the same $uw$. It might be hard to prove this via closure properties. On the other hand, it is quite straight-forward via automata.
$endgroup$
– Peter Leupold
Jan 28 at 12:44
$begingroup$
The $u$ are from the language right quotient of ($L_2$ by $L_1$) intersected with $L_1$; the $v$ are from $L_1$. But if you concatenate the three components, you cannot guarantee that the $u$ and the $w$ are from the same $uw$. It might be hard to prove this via closure properties. On the other hand, it is quite straight-forward via automata.
$endgroup$
– Peter Leupold
Jan 28 at 12:44
add a comment |
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$begingroup$
The $u$ are from the language right quotient of ($L_2$ by $L_1$) intersected with $L_1$; the $v$ are from $L_1$. But if you concatenate the three components, you cannot guarantee that the $u$ and the $w$ are from the same $uw$. It might be hard to prove this via closure properties. On the other hand, it is quite straight-forward via automata.
$endgroup$
– Peter Leupold
Jan 28 at 12:44