How to prove that $L={win{0,1,2}^*|#_2(w)<#_0(w)}$ is not regular using Myhill–Nerode theorem?
How to prove that $L={win{0,1,2}^*|#_2(w)<#_0(w)}$ is not regular using Myhill–Nerode theorem?
$#_2(w)$ means the number of occurrences of $2$ in $w$, same goes for $#_0(w)$.
I thought of the following proof:
Suppose that $L$ is regular then by Myhill–Nerode theorem the number of equivalence classes for $R_L$ is finite. ($R_L$ is a relation such that $xR_Lyiff forall zin Sigma^*: xzin Liff yzin L$).
Let us inspect words of type $0^i2^j, i>j$. Because there're infinitely many words in $L$ then according to pigeonhole principle some equivalence class must contain at least $2$ words. Then exist $i,j$ such that $i=j+1$ and $p,q$ such that $p=q+2$, $p>i$, $0^i2^jR_L0^p2^q$.
Then for $z=2^{j}$, $0^i2^jnotin L$ but $0^p2^qin L$ in contradiction to Myhill–Nerode theorem.
I'm wondering if it's OK to suppose quite concrete details like $i=j+1$ and $p=q+2, p>i$?
proof-verification formal-languages regular-language
edited Nov 24 at 14:23
Wuestenfux
3,1301410
asked Nov 24 at 14:02
Yos
1,127723
add a comment |
How to prove that $L={win{0,1,2}^*|#_2(w)<#_0(w)}$ is not regular using Myhill–Nerode theorem?
$#_2(w)$ means the number of occurrences of $2$ in $w$, same goes for $#_0(w)$.
I thought of the following proof:
Suppose that $L$ is regular then by Myhill–Nerode theorem the number of equivalence classes for $R_L$ is finite. ($R_L$ is a relation such that $xR_Lyiff forall zin Sigma^*: xzin Liff yzin L$).
Let us inspect words of type $0^i2^j, i>j$. Because there're infinitely many words in $L$ then according to pigeonhole principle some equivalence class must contain at least $2$ words. Then exist $i,j$ such that $i=j+1$ and $p,q$ such that $p=q+2$, $p>i$, $0^i2^jR_L0^p2^q$.
Then for $z=2^{j}$, $0^i2^jnotin L$ but $0^p2^qin L$ in contradiction to Myhill–Nerode theorem.
I'm wondering if it's OK to suppose quite concrete details like $i=j+1$ and $p=q+2, p>i$?
proof-verification formal-languages regular-language
edited Nov 24 at 14:23
Wuestenfux
3,1301410
asked Nov 24 at 14:02
Yos
1,127723
add a comment |
How to prove that $L={win{0,1,2}^*|#_2(w)<#_0(w)}$ is not regular using Myhill–Nerode theorem?
$#_2(w)$ means the number of occurrences of $2$ in $w$, same goes for $#_0(w)$.
I thought of the following proof:
Suppose that $L$ is regular then by Myhill–Nerode theorem the number of equivalence classes for $R_L$ is finite. ($R_L$ is a relation such that $xR_Lyiff forall zin Sigma^*: xzin Liff yzin L$).
Let us inspect words of type $0^i2^j, i>j$. Because there're infinitely many words in $L$ then according to pigeonhole principle some equivalence class must contain at least $2$ words. Then exist $i,j$ such that $i=j+1$ and $p,q$ such that $p=q+2$, $p>i$, $0^i2^jR_L0^p2^q$.
Then for $z=2^{j}$, $0^i2^jnotin L$ but $0^p2^qin L$ in contradiction to Myhill–Nerode theorem.
I'm wondering if it's OK to suppose quite concrete details like $i=j+1$ and $p=q+2, p>i$?
proof-verification formal-languages regular-language
edited Nov 24 at 14:23
Wuestenfux
3,1301410
asked Nov 24 at 14:02
Yos
1,127723
How to prove that $L={win{0,1,2}^*|#_2(w)<#_0(w)}$ is not regular using Myhill–Nerode theorem?
$#_2(w)$ means the number of occurrences of $2$ in $w$, same goes for $#_0(w)$.
I thought of the following proof:
Suppose that $L$ is regular then by Myhill–Nerode theorem the number of equivalence classes for $R_L$ is finite. ($R_L$ is a relation such that $xR_Lyiff forall zin Sigma^*: xzin Liff yzin L$).
Let us inspect words of type $0^i2^j, i>j$. Because there're infinitely many words in $L$ then according to pigeonhole principle some equivalence class must contain at least $2$ words. Then exist $i,j$ such that $i=j+1$ and $p,q$ such that $p=q+2$, $p>i$, $0^i2^jR_L0^p2^q$.
Then for $z=2^{j}$, $0^i2^jnotin L$ but $0^p2^qin L$ in contradiction to Myhill–Nerode theorem.
I'm wondering if it's OK to suppose quite concrete details like $i=j+1$ and $p=q+2, p>i$?
proof-verification formal-languages regular-language
proof-verification formal-languages regular-language
edited Nov 24 at 14:23
Wuestenfux
3,1301410
asked Nov 24 at 14:02
Yos
1,127723
edited Nov 24 at 14:23
Wuestenfux
3,1301410
asked Nov 24 at 14:02
Yos
1,127723
edited Nov 24 at 14:23
Wuestenfux
3,1301410
edited Nov 24 at 14:23
Wuestenfux
3,1301410
edited Nov 24 at 14:23
Wuestenfux
3,1301410
3,1301410
asked Nov 24 at 14:02
Yos
1,127723
asked Nov 24 at 14:02
Yos
1,127723
asked Nov 24 at 14:02
Yos
1,127723
1,127723
add a comment |
add a comment |
1 Answer
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active
oldest
votes
Hint: The words $0^k2^l$ and $0^k1^l$ with $l<k$ cannot be in relation $R_L$, since $0^k2^l(2^{k-l})=0^k2^knotin L$, but $0^k 1^l 2^{k-l}in L$ for $z=2^{k-l}$.
answered Nov 24 at 14:17
Wuestenfux
3,1301410
You're offering your solution but is my solution correct?
– Yos
Nov 24 at 14:20
Not sure about your contraints about $i,j$ and $p,q$.
– Wuestenfux
Nov 24 at 14:22
I appreciate your answer but I'd like to know whether my proof is OK and if not why.
– Yos
Nov 24 at 14:23
It would be correct if you would you would relieve the conditions about $i,j$, $p,q$.
– Wuestenfux
Nov 24 at 14:25
1
Just a word that falsifies equivalence $Leftrightarrow$.
– Wuestenfux
Nov 24 at 14:32
|
show 1 more comment
Your Answer
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1 Answer
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oldest
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Hint: The words $0^k2^l$ and $0^k1^l$ with $l<k$ cannot be in relation $R_L$, since $0^k2^l(2^{k-l})=0^k2^knotin L$, but $0^k 1^l 2^{k-l}in L$ for $z=2^{k-l}$.
answered Nov 24 at 14:17
Wuestenfux
3,1301410
You're offering your solution but is my solution correct?
– Yos
Nov 24 at 14:20
Not sure about your contraints about $i,j$ and $p,q$.
– Wuestenfux
Nov 24 at 14:22
I appreciate your answer but I'd like to know whether my proof is OK and if not why.
– Yos
Nov 24 at 14:23
It would be correct if you would you would relieve the conditions about $i,j$, $p,q$.
– Wuestenfux
Nov 24 at 14:25
1
Just a word that falsifies equivalence $Leftrightarrow$.
– Wuestenfux
Nov 24 at 14:32
|
show 1 more comment
Hint: The words $0^k2^l$ and $0^k1^l$ with $l<k$ cannot be in relation $R_L$, since $0^k2^l(2^{k-l})=0^k2^knotin L$, but $0^k 1^l 2^{k-l}in L$ for $z=2^{k-l}$.
answered Nov 24 at 14:17
Wuestenfux
3,1301410
You're offering your solution but is my solution correct?
– Yos
Nov 24 at 14:20
Not sure about your contraints about $i,j$ and $p,q$.
– Wuestenfux
Nov 24 at 14:22
I appreciate your answer but I'd like to know whether my proof is OK and if not why.
– Yos
Nov 24 at 14:23
It would be correct if you would you would relieve the conditions about $i,j$, $p,q$.
– Wuestenfux
Nov 24 at 14:25
1
Just a word that falsifies equivalence $Leftrightarrow$.
– Wuestenfux
Nov 24 at 14:32
|
show 1 more comment
Hint: The words $0^k2^l$ and $0^k1^l$ with $l<k$ cannot be in relation $R_L$, since $0^k2^l(2^{k-l})=0^k2^knotin L$, but $0^k 1^l 2^{k-l}in L$ for $z=2^{k-l}$.
answered Nov 24 at 14:17
Wuestenfux
3,1301410
Hint: The words $0^k2^l$ and $0^k1^l$ with $l<k$ cannot be in relation $R_L$, since $0^k2^l(2^{k-l})=0^k2^knotin L$, but $0^k 1^l 2^{k-l}in L$ for $z=2^{k-l}$.
answered Nov 24 at 14:17
Wuestenfux
3,1301410
answered Nov 24 at 14:17
Wuestenfux
3,1301410
answered Nov 24 at 14:17
Wuestenfux
3,1301410
answered Nov 24 at 14:17
Wuestenfux
3,1301410
3,1301410
You're offering your solution but is my solution correct?
– Yos
Nov 24 at 14:20
Not sure about your contraints about $i,j$ and $p,q$.
– Wuestenfux
Nov 24 at 14:22
I appreciate your answer but I'd like to know whether my proof is OK and if not why.
– Yos
Nov 24 at 14:23
It would be correct if you would you would relieve the conditions about $i,j$, $p,q$.
– Wuestenfux
Nov 24 at 14:25
1
Just a word that falsifies equivalence $Leftrightarrow$.
– Wuestenfux
Nov 24 at 14:32
|
show 1 more comment
You're offering your solution but is my solution correct?
– Yos
Nov 24 at 14:20
Not sure about your contraints about $i,j$ and $p,q$.
– Wuestenfux
Nov 24 at 14:22
I appreciate your answer but I'd like to know whether my proof is OK and if not why.
– Yos
Nov 24 at 14:23
It would be correct if you would you would relieve the conditions about $i,j$, $p,q$.
– Wuestenfux
Nov 24 at 14:25
1
Just a word that falsifies equivalence $Leftrightarrow$.
– Wuestenfux
Nov 24 at 14:32
You're offering your solution but is my solution correct?
– Yos
Nov 24 at 14:20
You're offering your solution but is my solution correct?
– Yos
Nov 24 at 14:20
Not sure about your contraints about $i,j$ and $p,q$.
– Wuestenfux
Nov 24 at 14:22
Not sure about your contraints about $i,j$ and $p,q$.
– Wuestenfux
Nov 24 at 14:22
I appreciate your answer but I'd like to know whether my proof is OK and if not why.
– Yos
Nov 24 at 14:23
I appreciate your answer but I'd like to know whether my proof is OK and if not why.
– Yos
Nov 24 at 14:23
It would be correct if you would you would relieve the conditions about $i,j$, $p,q$.
– Wuestenfux
Nov 24 at 14:25
It would be correct if you would you would relieve the conditions about $i,j$, $p,q$.
– Wuestenfux
Nov 24 at 14:25
1
1
Just a word that falsifies equivalence $Leftrightarrow$.
– Wuestenfux
Nov 24 at 14:32
Just a word that falsifies equivalence $Leftrightarrow$.
– Wuestenfux
Nov 24 at 14:32
|
show 1 more comment
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