Transforming an English Sentence Into Predicate Logic












0












$begingroup$



"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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$



"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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$



"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







share|cite|improve this question











$endgroup$





"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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 8:59









Mauro ALLEGRANZA

66.8k449115




66.8k449115










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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 $C(everybody)$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
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active

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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 $C(everybody)$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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 $C(everybody)$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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 $C(everybody)$?






share|cite|improve this answer









$endgroup$



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 $C(everybody)$?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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