Transforming an English Sentence Into Predicate Logic
$begingroup$
"Not Everybody likes everybody."
I have the problem that everybody is both the subject and object of that sentence.
I am not sure if that is correct: $exists x lnot text l(x, x)$,
where x = everybody and $text l(x, x) =$ "x likes x".
UPDATE 1: Second attempt
$lnot forall y in x forall z in x text l(y,z)$
UPDATE 2:
There exists some person y that does not like every z:
$exists y lnot forall z text l(y,z)$
logic predicate-logic
edited Dec 14 '18 at 8:59
Mauro ALLEGRANZA
66.8k449115
asked Dec 13 '18 at 23:15
WilliWilli
285
$endgroup$
add a comment |
$begingroup$
"Not Everybody likes everybody."
I have the problem that everybody is both the subject and object of that sentence.
I am not sure if that is correct: $exists x lnot text l(x, x)$,
where x = everybody and $text l(x, x) =$ "x likes x".
UPDATE 1: Second attempt
$lnot forall y in x forall z in x text l(y,z)$
UPDATE 2:
There exists some person y that does not like every z:
$exists y lnot forall z text l(y,z)$
logic predicate-logic
edited Dec 14 '18 at 8:59
Mauro ALLEGRANZA
66.8k449115
asked Dec 13 '18 at 23:15
WilliWilli
285
$endgroup$
$begingroup$
I'd define l(x,y) as x likes y, but otherwise this is correct.
$endgroup$
– user458276
Dec 13 '18 at 23:27
$begingroup$
"everybody likes everybody" is "for all people x, for all people y, x likes y:. Now negate that.un English first. It begins "there is some person x ..."
$endgroup$
– Ethan Bolker
Dec 13 '18 at 23:57
add a comment |
$begingroup$
"Not Everybody likes everybody."
I have the problem that everybody is both the subject and object of that sentence.
I am not sure if that is correct: $exists x lnot text l(x, x)$,
where x = everybody and $text l(x, x) =$ "x likes x".
UPDATE 1: Second attempt
$lnot forall y in x forall z in x text l(y,z)$
UPDATE 2:
There exists some person y that does not like every z:
$exists y lnot forall z text l(y,z)$
logic predicate-logic
edited Dec 14 '18 at 8:59
Mauro ALLEGRANZA
66.8k449115
asked Dec 13 '18 at 23:15
WilliWilli
285
$endgroup$
"Not Everybody likes everybody."
I have the problem that everybody is both the subject and object of that sentence.
I am not sure if that is correct: $exists x lnot text l(x, x)$,
where x = everybody and $text l(x, x) =$ "x likes x".
UPDATE 1: Second attempt
$lnot forall y in x forall z in x text l(y,z)$
UPDATE 2:
There exists some person y that does not like every z:
$exists y lnot forall z text l(y,z)$
logic predicate-logic
logic predicate-logic
edited Dec 14 '18 at 8:59
Mauro ALLEGRANZA
66.8k449115
asked Dec 13 '18 at 23:15
WilliWilli
285
edited Dec 14 '18 at 8:59
Mauro ALLEGRANZA
66.8k449115
asked Dec 13 '18 at 23:15
WilliWilli
285
edited Dec 14 '18 at 8:59
Mauro ALLEGRANZA
66.8k449115
edited Dec 14 '18 at 8:59
Mauro ALLEGRANZA
66.8k449115
edited Dec 14 '18 at 8:59
Mauro ALLEGRANZA
66.8k449115
66.8k449115
asked Dec 13 '18 at 23:15
WilliWilli
285
asked Dec 13 '18 at 23:15
WilliWilli
285
asked Dec 13 '18 at 23:15
WilliWilli
285
285
$begingroup$
I'd define l(x,y) as x likes y, but otherwise this is correct.
$endgroup$
– user458276
Dec 13 '18 at 23:27
$begingroup$
"everybody likes everybody" is "for all people x, for all people y, x likes y:. Now negate that.un English first. It begins "there is some person x ..."
$endgroup$
– Ethan Bolker
Dec 13 '18 at 23:57
add a comment |
$begingroup$
I'd define l(x,y) as x likes y, but otherwise this is correct.
$endgroup$
– user458276
Dec 13 '18 at 23:27
$begingroup$
"everybody likes everybody" is "for all people x, for all people y, x likes y:. Now negate that.un English first. It begins "there is some person x ..."
$endgroup$
– Ethan Bolker
Dec 13 '18 at 23:57
$begingroup$
I'd define l(x,y) as x likes y, but otherwise this is correct.
$endgroup$
– user458276
Dec 13 '18 at 23:27
$begingroup$
I'd define l(x,y) as x likes y, but otherwise this is correct.
$endgroup$
– user458276
Dec 13 '18 at 23:27
$begingroup$
"everybody likes everybody" is "for all people x, for all people y, x likes y:. Now negate that.un English first. It begins "there is some person x ..."
$endgroup$
– Ethan Bolker
Dec 13 '18 at 23:57
$begingroup$
"everybody likes everybody" is "for all people x, for all people y, x likes y:. Now negate that.un English first. It begins "there is some person x ..."
$endgroup$
– Ethan Bolker
Dec 13 '18 at 23:57
add a comment |
1 Answer
1
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votes
$begingroup$
I disagree with user458276’s comment. You can’t say what $x$ is at the same time you use it as the variable for $exists$. The variable after $exists$ is “bound” and represents a particular but arbitrary element of a set (which you didn’t identify). For $x$ to take on a specific value, it must be “free” and able to be substituted with a value. You are also avoiding dealing with what “everybody” means by trying to let it be the value of a variable. If you had to write “Everybody likes cheese.” in propositional logic, would you let $C(x)$ mean “$x$ likes cheese” and write the sentence as $C(everybody)$?
answered Dec 13 '18 at 23:32
Steve KassSteve Kass
11.4k11530
$endgroup$
$begingroup$
I made an update to my question inspired by your comment.
$endgroup$
– Willi
Dec 13 '18 at 23:55
1
$begingroup$
Your update is good! Note: You could also move the $lnot$ through the $forall$ to get a version with only $exists$, and it would also be helpful to specify the set (of all people) that is the domain of quantification here.
$endgroup$
– Steve Kass
Dec 14 '18 at 1:51
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I disagree with user458276’s comment. You can’t say what $x$ is at the same time you use it as the variable for $exists$. The variable after $exists$ is “bound” and represents a particular but arbitrary element of a set (which you didn’t identify). For $x$ to take on a specific value, it must be “free” and able to be substituted with a value. You are also avoiding dealing with what “everybody” means by trying to let it be the value of a variable. If you had to write “Everybody likes cheese.” in propositional logic, would you let $C(x)$ mean “$x$ likes cheese” and write the sentence as $C(everybody)$?
answered Dec 13 '18 at 23:32
Steve KassSteve Kass
11.4k11530
$endgroup$
$begingroup$
I made an update to my question inspired by your comment.
$endgroup$
– Willi
Dec 13 '18 at 23:55
1
$begingroup$
Your update is good! Note: You could also move the $lnot$ through the $forall$ to get a version with only $exists$, and it would also be helpful to specify the set (of all people) that is the domain of quantification here.
$endgroup$
– Steve Kass
Dec 14 '18 at 1:51
add a comment |
$begingroup$
I disagree with user458276’s comment. You can’t say what $x$ is at the same time you use it as the variable for $exists$. The variable after $exists$ is “bound” and represents a particular but arbitrary element of a set (which you didn’t identify). For $x$ to take on a specific value, it must be “free” and able to be substituted with a value. You are also avoiding dealing with what “everybody” means by trying to let it be the value of a variable. If you had to write “Everybody likes cheese.” in propositional logic, would you let $C(x)$ mean “$x$ likes cheese” and write the sentence as $C(everybody)$?
answered Dec 13 '18 at 23:32
Steve KassSteve Kass
11.4k11530
$endgroup$
$begingroup$
I made an update to my question inspired by your comment.
$endgroup$
– Willi
Dec 13 '18 at 23:55
1
$begingroup$
Your update is good! Note: You could also move the $lnot$ through the $forall$ to get a version with only $exists$, and it would also be helpful to specify the set (of all people) that is the domain of quantification here.
$endgroup$
– Steve Kass
Dec 14 '18 at 1:51
add a comment |
$begingroup$
I disagree with user458276’s comment. You can’t say what $x$ is at the same time you use it as the variable for $exists$. The variable after $exists$ is “bound” and represents a particular but arbitrary element of a set (which you didn’t identify). For $x$ to take on a specific value, it must be “free” and able to be substituted with a value. You are also avoiding dealing with what “everybody” means by trying to let it be the value of a variable. If you had to write “Everybody likes cheese.” in propositional logic, would you let $C(x)$ mean “$x$ likes cheese” and write the sentence as $C(everybody)$?
answered Dec 13 '18 at 23:32
Steve KassSteve Kass
11.4k11530
$endgroup$
I disagree with user458276’s comment. You can’t say what $x$ is at the same time you use it as the variable for $exists$. The variable after $exists$ is “bound” and represents a particular but arbitrary element of a set (which you didn’t identify). For $x$ to take on a specific value, it must be “free” and able to be substituted with a value. You are also avoiding dealing with what “everybody” means by trying to let it be the value of a variable. If you had to write “Everybody likes cheese.” in propositional logic, would you let $C(x)$ mean “$x$ likes cheese” and write the sentence as $C(everybody)$?
answered Dec 13 '18 at 23:32
Steve KassSteve Kass
11.4k11530
answered Dec 13 '18 at 23:32
Steve KassSteve Kass
11.4k11530
answered Dec 13 '18 at 23:32
Steve KassSteve Kass
11.4k11530
answered Dec 13 '18 at 23:32
Steve KassSteve Kass
11.4k11530
11.4k11530
$begingroup$
I made an update to my question inspired by your comment.
$endgroup$
– Willi
Dec 13 '18 at 23:55
1
$begingroup$
Your update is good! Note: You could also move the $lnot$ through the $forall$ to get a version with only $exists$, and it would also be helpful to specify the set (of all people) that is the domain of quantification here.
$endgroup$
– Steve Kass
Dec 14 '18 at 1:51
add a comment |
$begingroup$
I made an update to my question inspired by your comment.
$endgroup$
– Willi
Dec 13 '18 at 23:55
1
$begingroup$
Your update is good! Note: You could also move the $lnot$ through the $forall$ to get a version with only $exists$, and it would also be helpful to specify the set (of all people) that is the domain of quantification here.
$endgroup$
– Steve Kass
Dec 14 '18 at 1:51
$begingroup$
I made an update to my question inspired by your comment.
$endgroup$
– Willi
Dec 13 '18 at 23:55
$begingroup$
I made an update to my question inspired by your comment.
$endgroup$
– Willi
Dec 13 '18 at 23:55
1
1
$begingroup$
Your update is good! Note: You could also move the $lnot$ through the $forall$ to get a version with only $exists$, and it would also be helpful to specify the set (of all people) that is the domain of quantification here.
$endgroup$
– Steve Kass
Dec 14 '18 at 1:51
$begingroup$
Your update is good! Note: You could also move the $lnot$ through the $forall$ to get a version with only $exists$, and it would also be helpful to specify the set (of all people) that is the domain of quantification here.
$endgroup$
– Steve Kass
Dec 14 '18 at 1:51
add a comment |
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$begingroup$
I'd define l(x,y) as x likes y, but otherwise this is correct.
$endgroup$
– user458276
Dec 13 '18 at 23:27
$begingroup$
"everybody likes everybody" is "for all people x, for all people y, x likes y:. Now negate that.un English first. It begins "there is some person x ..."
$endgroup$
– Ethan Bolker
Dec 13 '18 at 23:57