What is the meaning of this ampersand and exclamation mark?
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On the top of page 4 of this paper Direct Construction of Minimal Acyclic Subsequential Transducers, there is an expression
$$forall r in S forall a in Sigma forall sigma in Sigma^* ((mu^*(s,sigma)=r & !mu(r,a)) to lambda(r,a)=[gmathcal T(sigma) ]^{-1}gmathcal T(sigma a))$$
What's the meaning of the ampersand and the exclamation mark? I guess that $&$ means "and", but shouldn't it be $land$?
Thanks.
notation
asked Jan 13 at 4:56
markericmarkeric
1
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add a comment |
$begingroup$
On the top of page 4 of this paper Direct Construction of Minimal Acyclic Subsequential Transducers, there is an expression
$$forall r in S forall a in Sigma forall sigma in Sigma^* ((mu^*(s,sigma)=r & !mu(r,a)) to lambda(r,a)=[gmathcal T(sigma) ]^{-1}gmathcal T(sigma a))$$
What's the meaning of the ampersand and the exclamation mark? I guess that $&$ means "and", but shouldn't it be $land$?
Thanks.
notation
asked Jan 13 at 4:56
markericmarkeric
1
$endgroup$
$begingroup$
I am not an expert on this topic, but I guess authors uses $&$ to denote 'and' because $land$ has another meaning.
$endgroup$
– Hanul Jeon
Jan 13 at 5:06
$begingroup$
$!$ may denotes the negation, but I don't know, I cannot say it with certainty.
$endgroup$
– Hanul Jeon
Jan 13 at 5:07
$begingroup$
I also though ! might mean negation, but $mu$ is not a boolean function so I'm not sure. In [4] they say $!mu(r,sigma)$ means $mu(r,sigma)$ has been defined.
$endgroup$
– tch
Jan 13 at 5:17
$begingroup$
@tch Thanks for point that out. I think you are right. The author borrowed the notation from that paper.
$endgroup$
– markeric
Jan 13 at 8:41
add a comment |
$begingroup$
On the top of page 4 of this paper Direct Construction of Minimal Acyclic Subsequential Transducers, there is an expression
$$forall r in S forall a in Sigma forall sigma in Sigma^* ((mu^*(s,sigma)=r & !mu(r,a)) to lambda(r,a)=[gmathcal T(sigma) ]^{-1}gmathcal T(sigma a))$$
What's the meaning of the ampersand and the exclamation mark? I guess that $&$ means "and", but shouldn't it be $land$?
Thanks.
notation
asked Jan 13 at 4:56
markericmarkeric
1
$endgroup$
On the top of page 4 of this paper Direct Construction of Minimal Acyclic Subsequential Transducers, there is an expression
$$forall r in S forall a in Sigma forall sigma in Sigma^* ((mu^*(s,sigma)=r & !mu(r,a)) to lambda(r,a)=[gmathcal T(sigma) ]^{-1}gmathcal T(sigma a))$$
What's the meaning of the ampersand and the exclamation mark? I guess that $&$ means "and", but shouldn't it be $land$?
Thanks.
notation
notation
asked Jan 13 at 4:56
markericmarkeric
1
asked Jan 13 at 4:56
markericmarkeric
1
asked Jan 13 at 4:56
markericmarkeric
1
asked Jan 13 at 4:56
markericmarkeric
1
asked Jan 13 at 4:56
markericmarkeric
1
1
$begingroup$
I am not an expert on this topic, but I guess authors uses $&$ to denote 'and' because $land$ has another meaning.
$endgroup$
– Hanul Jeon
Jan 13 at 5:06
$begingroup$
$!$ may denotes the negation, but I don't know, I cannot say it with certainty.
$endgroup$
– Hanul Jeon
Jan 13 at 5:07
$begingroup$
I also though ! might mean negation, but $mu$ is not a boolean function so I'm not sure. In [4] they say $!mu(r,sigma)$ means $mu(r,sigma)$ has been defined.
$endgroup$
– tch
Jan 13 at 5:17
$begingroup$
@tch Thanks for point that out. I think you are right. The author borrowed the notation from that paper.
$endgroup$
– markeric
Jan 13 at 8:41
add a comment |
$begingroup$
I am not an expert on this topic, but I guess authors uses $&$ to denote 'and' because $land$ has another meaning.
$endgroup$
– Hanul Jeon
Jan 13 at 5:06
$begingroup$
$!$ may denotes the negation, but I don't know, I cannot say it with certainty.
$endgroup$
– Hanul Jeon
Jan 13 at 5:07
$begingroup$
I also though ! might mean negation, but $mu$ is not a boolean function so I'm not sure. In [4] they say $!mu(r,sigma)$ means $mu(r,sigma)$ has been defined.
$endgroup$
– tch
Jan 13 at 5:17
$begingroup$
@tch Thanks for point that out. I think you are right. The author borrowed the notation from that paper.
$endgroup$
– markeric
Jan 13 at 8:41
$begingroup$
I am not an expert on this topic, but I guess authors uses $&$ to denote 'and' because $land$ has another meaning.
$endgroup$
– Hanul Jeon
Jan 13 at 5:06
$begingroup$
I am not an expert on this topic, but I guess authors uses $&$ to denote 'and' because $land$ has another meaning.
$endgroup$
– Hanul Jeon
Jan 13 at 5:06
$begingroup$
$!$ may denotes the negation, but I don't know, I cannot say it with certainty.
$endgroup$
– Hanul Jeon
Jan 13 at 5:07
$begingroup$
$!$ may denotes the negation, but I don't know, I cannot say it with certainty.
$endgroup$
– Hanul Jeon
Jan 13 at 5:07
$begingroup$
I also though ! might mean negation, but $mu$ is not a boolean function so I'm not sure. In [4] they say $!mu(r,sigma)$ means $mu(r,sigma)$ has been defined.
$endgroup$
– tch
Jan 13 at 5:17
$begingroup$
I also though ! might mean negation, but $mu$ is not a boolean function so I'm not sure. In [4] they say $!mu(r,sigma)$ means $mu(r,sigma)$ has been defined.
$endgroup$
– tch
Jan 13 at 5:17
$begingroup$
@tch Thanks for point that out. I think you are right. The author borrowed the notation from that paper.
$endgroup$
– markeric
Jan 13 at 8:41
$begingroup$
@tch Thanks for point that out. I think you are right. The author borrowed the notation from that paper.
$endgroup$
– markeric
Jan 13 at 8:41
add a comment |
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$begingroup$
I am not an expert on this topic, but I guess authors uses $&$ to denote 'and' because $land$ has another meaning.
$endgroup$
– Hanul Jeon
Jan 13 at 5:06
$begingroup$
$!$ may denotes the negation, but I don't know, I cannot say it with certainty.
$endgroup$
– Hanul Jeon
Jan 13 at 5:07
$begingroup$
I also though ! might mean negation, but $mu$ is not a boolean function so I'm not sure. In [4] they say $!mu(r,sigma)$ means $mu(r,sigma)$ has been defined.
$endgroup$
– tch
Jan 13 at 5:17
$begingroup$
@tch Thanks for point that out. I think you are right. The author borrowed the notation from that paper.
$endgroup$
– markeric
Jan 13 at 8:41