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Translate from English to first-order logic: There is somebody who likes everyone who likes John
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Translate from English to first-order logic: There is somebody who likes everyone who likes John
I'm not sure about the answer. What I'm thinking is $(∃x)(L(x,y)∧L(x,J))$
discrete-mathematics logic logic-translation
edited Aug 6 at 16:51
Bram28
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asked Apr 27 '17 at 2:00
Haley_Q
364
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Translate from English to first-order logic: There is somebody who likes everyone who likes John
I'm not sure about the answer. What I'm thinking is $(∃x)(L(x,y)∧L(x,J))$
discrete-mathematics logic logic-translation
edited Aug 6 at 16:51
Bram28
55.2k33982
asked Apr 27 '17 at 2:00
Haley_Q
364
I'm assuming $L(x,y)$ means $x$ likes $y$. Reading your thought to myself, I read: "There is somebody ($x$) who likes $y$ and who likes $J$."
– Mr Toad
Apr 27 '17 at 2:07
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Translate from English to first-order logic: There is somebody who likes everyone who likes John
I'm not sure about the answer. What I'm thinking is $(∃x)(L(x,y)∧L(x,J))$
discrete-mathematics logic logic-translation
edited Aug 6 at 16:51
Bram28
55.2k33982
asked Apr 27 '17 at 2:00
Haley_Q
364
Translate from English to first-order logic: There is somebody who likes everyone who likes John
I'm not sure about the answer. What I'm thinking is $(∃x)(L(x,y)∧L(x,J))$
discrete-mathematics logic logic-translation
edited Aug 6 at 16:51
Bram28
55.2k33982
asked Apr 27 '17 at 2:00
Haley_Q
364
edited Aug 6 at 16:51
Bram28
55.2k33982
edited Aug 6 at 16:51
Bram28
55.2k33982
edited Aug 6 at 16:51
Bram28
55.2k33982
55.2k33982
asked Apr 27 '17 at 2:00
Haley_Q
364
asked Apr 27 '17 at 2:00
Haley_Q
364
asked Apr 27 '17 at 2:00
Haley_Q
364
364
I'm assuming $L(x,y)$ means $x$ likes $y$. Reading your thought to myself, I read: "There is somebody ($x$) who likes $y$ and who likes $J$."
– Mr Toad
Apr 27 '17 at 2:07
add a comment |Â
I'm assuming $L(x,y)$ means $x$ likes $y$. Reading your thought to myself, I read: "There is somebody ($x$) who likes $y$ and who likes $J$."
– Mr Toad
Apr 27 '17 at 2:07
I'm assuming $L(x,y)$ means $x$ likes $y$. Reading your thought to myself, I read: "There is somebody ($x$) who likes $y$ and who likes $J$."
– Mr Toad
Apr 27 '17 at 2:07
I'm assuming $L(x,y)$ means $x$ likes $y$. Reading your thought to myself, I read: "There is somebody ($x$) who likes $y$ and who likes $J$."
– Mr Toad
Apr 27 '17 at 2:07
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
$$(exists x)(forall y)(L(y,J)rightarrow L(x,y))$$
Translation:
There exists $x$ such that if $y$ likes John, then $x$ likes $y$.
answered Apr 27 '17 at 2:07
quasi
33.4k22359
Don't you mean $(exists x)(forall y)(L(y,J)to L(x,y))$? The way you wrote it, $y$ is a free variable.
– bof
Apr 29 '17 at 5:16
I guess I assumed that an unquantified variable not specified as constant was automatically quantfified, by default, with "for all". But I'll edit my answer to make that explicit. Thanks.
– quasi
Apr 29 '17 at 5:24
Wouldn't the default be $(forall y)(exists x)(L(y,J)to L(x,y))$?
– bof
Apr 29 '17 at 5:29
I guess so, so the edit was necessary, else the default order would yield the wrong meaning. Thanks again.
– quasi
Apr 29 '17 at 5:41
add a comment |Â
up vote
1
down vote
There is some person who likes everyone who likes John. So, there is some person $x$ who likes all people $y$ who satisfy the condition that $y$ likes John. So, there exists a person $x$ such that for all $y$, if $y$ likes John then $x$ likes $y$. We have the expression: $exists x, forall y (L(y,J) implies L(x,y))$.
answered Apr 29 '17 at 5:12
Ashwin Ganesan
3,21069
add a comment |Â
up vote
1
down vote
The English language is not logically precise, so the sentence has two (possible more) interpretations. The first, as given by @quasi, is "there is somebody who likes (everybody who likes John)", and has the translation $(exists x)(forall y)(L(y,J)rightarrow L(x,y))$. The second is "there is (somebody who likes everybody) who likes John", which has the translation $(exists x)[(forall y(L(x,y))),land,L(x,J)]$. Here $L(a,b)$ means $a$ likes $b$.
(The first is probably the intended meaning, because if somebody liked everyone, they would surely like John. I'm just pointing out an interpretation closer to what you were originally thinking.)
answered Apr 27 '17 at 2:52
shardulc
3,620927
Thank you. The second answer is what I originally thinking. I was not sure how to translate it correctly. I agree with you that the first translation might be the intended meaning.
– Haley_Q
Apr 27 '17 at 3:14
@Haley_Q You're welcome! If you like my answer, please upvote it (the up arrow) and accept it if it answers your question.
– shardulc
Apr 27 '17 at 3:31
For the first interpretation, do you really want $y$ to be a free variable, or did you miss out a $forall y$?
– bof
Apr 29 '17 at 5:17
@bof Thanks, fixed it.
– shardulc
Apr 29 '17 at 5:56
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$$(exists x)(forall y)(L(y,J)rightarrow L(x,y))$$
Translation:
There exists $x$ such that if $y$ likes John, then $x$ likes $y$.
answered Apr 27 '17 at 2:07
quasi
33.4k22359
Don't you mean $(exists x)(forall y)(L(y,J)to L(x,y))$? The way you wrote it, $y$ is a free variable.
– bof
Apr 29 '17 at 5:16
I guess I assumed that an unquantified variable not specified as constant was automatically quantfified, by default, with "for all". But I'll edit my answer to make that explicit. Thanks.
– quasi
Apr 29 '17 at 5:24
Wouldn't the default be $(forall y)(exists x)(L(y,J)to L(x,y))$?
– bof
Apr 29 '17 at 5:29
I guess so, so the edit was necessary, else the default order would yield the wrong meaning. Thanks again.
– quasi
Apr 29 '17 at 5:41
add a comment |Â
up vote
2
down vote
$$(exists x)(forall y)(L(y,J)rightarrow L(x,y))$$
Translation:
There exists $x$ such that if $y$ likes John, then $x$ likes $y$.
answered Apr 27 '17 at 2:07
quasi
33.4k22359
Don't you mean $(exists x)(forall y)(L(y,J)to L(x,y))$? The way you wrote it, $y$ is a free variable.
– bof
Apr 29 '17 at 5:16
I guess I assumed that an unquantified variable not specified as constant was automatically quantfified, by default, with "for all". But I'll edit my answer to make that explicit. Thanks.
– quasi
Apr 29 '17 at 5:24
Wouldn't the default be $(forall y)(exists x)(L(y,J)to L(x,y))$?
– bof
Apr 29 '17 at 5:29
I guess so, so the edit was necessary, else the default order would yield the wrong meaning. Thanks again.
– quasi
Apr 29 '17 at 5:41
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$$(exists x)(forall y)(L(y,J)rightarrow L(x,y))$$
Translation:
There exists $x$ such that if $y$ likes John, then $x$ likes $y$.
answered Apr 27 '17 at 2:07
quasi
33.4k22359
$$(exists x)(forall y)(L(y,J)rightarrow L(x,y))$$
Translation:
There exists $x$ such that if $y$ likes John, then $x$ likes $y$.
answered Apr 27 '17 at 2:07
quasi
33.4k22359
edited Apr 29 '17 at 5:25
answered Apr 27 '17 at 2:07
quasi
33.4k22359
answered Apr 27 '17 at 2:07
quasi
33.4k22359
answered Apr 27 '17 at 2:07
quasi
33.4k22359
33.4k22359
Don't you mean $(exists x)(forall y)(L(y,J)to L(x,y))$? The way you wrote it, $y$ is a free variable.
– bof
Apr 29 '17 at 5:16
I guess I assumed that an unquantified variable not specified as constant was automatically quantfified, by default, with "for all". But I'll edit my answer to make that explicit. Thanks.
– quasi
Apr 29 '17 at 5:24
Wouldn't the default be $(forall y)(exists x)(L(y,J)to L(x,y))$?
– bof
Apr 29 '17 at 5:29
I guess so, so the edit was necessary, else the default order would yield the wrong meaning. Thanks again.
– quasi
Apr 29 '17 at 5:41
add a comment |Â
Don't you mean $(exists x)(forall y)(L(y,J)to L(x,y))$? The way you wrote it, $y$ is a free variable.
– bof
Apr 29 '17 at 5:16
I guess I assumed that an unquantified variable not specified as constant was automatically quantfified, by default, with "for all". But I'll edit my answer to make that explicit. Thanks.
– quasi
Apr 29 '17 at 5:24
Wouldn't the default be $(forall y)(exists x)(L(y,J)to L(x,y))$?
– bof
Apr 29 '17 at 5:29
I guess so, so the edit was necessary, else the default order would yield the wrong meaning. Thanks again.
– quasi
Apr 29 '17 at 5:41
Don't you mean $(exists x)(forall y)(L(y,J)to L(x,y))$? The way you wrote it, $y$ is a free variable.
– bof
Apr 29 '17 at 5:16
Don't you mean $(exists x)(forall y)(L(y,J)to L(x,y))$? The way you wrote it, $y$ is a free variable.
– bof
Apr 29 '17 at 5:16
I guess I assumed that an unquantified variable not specified as constant was automatically quantfified, by default, with "for all". But I'll edit my answer to make that explicit. Thanks.
– quasi
Apr 29 '17 at 5:24
I guess I assumed that an unquantified variable not specified as constant was automatically quantfified, by default, with "for all". But I'll edit my answer to make that explicit. Thanks.
– quasi
Apr 29 '17 at 5:24
Wouldn't the default be $(forall y)(exists x)(L(y,J)to L(x,y))$?
– bof
Apr 29 '17 at 5:29
Wouldn't the default be $(forall y)(exists x)(L(y,J)to L(x,y))$?
– bof
Apr 29 '17 at 5:29
I guess so, so the edit was necessary, else the default order would yield the wrong meaning. Thanks again.
– quasi
Apr 29 '17 at 5:41
I guess so, so the edit was necessary, else the default order would yield the wrong meaning. Thanks again.
– quasi
Apr 29 '17 at 5:41
add a comment |Â
up vote
1
down vote
There is some person who likes everyone who likes John. So, there is some person $x$ who likes all people $y$ who satisfy the condition that $y$ likes John. So, there exists a person $x$ such that for all $y$, if $y$ likes John then $x$ likes $y$. We have the expression: $exists x, forall y (L(y,J) implies L(x,y))$.
answered Apr 29 '17 at 5:12
Ashwin Ganesan
3,21069
add a comment |Â
up vote
1
down vote
There is some person who likes everyone who likes John. So, there is some person $x$ who likes all people $y$ who satisfy the condition that $y$ likes John. So, there exists a person $x$ such that for all $y$, if $y$ likes John then $x$ likes $y$. We have the expression: $exists x, forall y (L(y,J) implies L(x,y))$.
answered Apr 29 '17 at 5:12
Ashwin Ganesan
3,21069
add a comment |Â
up vote
1
down vote
up vote
1
down vote
There is some person who likes everyone who likes John. So, there is some person $x$ who likes all people $y$ who satisfy the condition that $y$ likes John. So, there exists a person $x$ such that for all $y$, if $y$ likes John then $x$ likes $y$. We have the expression: $exists x, forall y (L(y,J) implies L(x,y))$.
answered Apr 29 '17 at 5:12
Ashwin Ganesan
3,21069
There is some person who likes everyone who likes John. So, there is some person $x$ who likes all people $y$ who satisfy the condition that $y$ likes John. So, there exists a person $x$ such that for all $y$, if $y$ likes John then $x$ likes $y$. We have the expression: $exists x, forall y (L(y,J) implies L(x,y))$.
answered Apr 29 '17 at 5:12
Ashwin Ganesan
3,21069
answered Apr 29 '17 at 5:12
Ashwin Ganesan
3,21069
answered Apr 29 '17 at 5:12
Ashwin Ganesan
3,21069
answered Apr 29 '17 at 5:12
Ashwin Ganesan
3,21069
3,21069
add a comment |Â
add a comment |Â
up vote
1
down vote
The English language is not logically precise, so the sentence has two (possible more) interpretations. The first, as given by @quasi, is "there is somebody who likes (everybody who likes John)", and has the translation $(exists x)(forall y)(L(y,J)rightarrow L(x,y))$. The second is "there is (somebody who likes everybody) who likes John", which has the translation $(exists x)[(forall y(L(x,y))),land,L(x,J)]$. Here $L(a,b)$ means $a$ likes $b$.
(The first is probably the intended meaning, because if somebody liked everyone, they would surely like John. I'm just pointing out an interpretation closer to what you were originally thinking.)
answered Apr 27 '17 at 2:52
shardulc
3,620927
Thank you. The second answer is what I originally thinking. I was not sure how to translate it correctly. I agree with you that the first translation might be the intended meaning.
– Haley_Q
Apr 27 '17 at 3:14
@Haley_Q You're welcome! If you like my answer, please upvote it (the up arrow) and accept it if it answers your question.
– shardulc
Apr 27 '17 at 3:31
For the first interpretation, do you really want $y$ to be a free variable, or did you miss out a $forall y$?
– bof
Apr 29 '17 at 5:17
@bof Thanks, fixed it.
– shardulc
Apr 29 '17 at 5:56
add a comment |Â
up vote
1
down vote
The English language is not logically precise, so the sentence has two (possible more) interpretations. The first, as given by @quasi, is "there is somebody who likes (everybody who likes John)", and has the translation $(exists x)(forall y)(L(y,J)rightarrow L(x,y))$. The second is "there is (somebody who likes everybody) who likes John", which has the translation $(exists x)[(forall y(L(x,y))),land,L(x,J)]$. Here $L(a,b)$ means $a$ likes $b$.
(The first is probably the intended meaning, because if somebody liked everyone, they would surely like John. I'm just pointing out an interpretation closer to what you were originally thinking.)
answered Apr 27 '17 at 2:52
shardulc
3,620927
Thank you. The second answer is what I originally thinking. I was not sure how to translate it correctly. I agree with you that the first translation might be the intended meaning.
– Haley_Q
Apr 27 '17 at 3:14
@Haley_Q You're welcome! If you like my answer, please upvote it (the up arrow) and accept it if it answers your question.
– shardulc
Apr 27 '17 at 3:31
For the first interpretation, do you really want $y$ to be a free variable, or did you miss out a $forall y$?
– bof
Apr 29 '17 at 5:17
@bof Thanks, fixed it.
– shardulc
Apr 29 '17 at 5:56
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The English language is not logically precise, so the sentence has two (possible more) interpretations. The first, as given by @quasi, is "there is somebody who likes (everybody who likes John)", and has the translation $(exists x)(forall y)(L(y,J)rightarrow L(x,y))$. The second is "there is (somebody who likes everybody) who likes John", which has the translation $(exists x)[(forall y(L(x,y))),land,L(x,J)]$. Here $L(a,b)$ means $a$ likes $b$.
(The first is probably the intended meaning, because if somebody liked everyone, they would surely like John. I'm just pointing out an interpretation closer to what you were originally thinking.)
answered Apr 27 '17 at 2:52
shardulc
3,620927
The English language is not logically precise, so the sentence has two (possible more) interpretations. The first, as given by @quasi, is "there is somebody who likes (everybody who likes John)", and has the translation $(exists x)(forall y)(L(y,J)rightarrow L(x,y))$. The second is "there is (somebody who likes everybody) who likes John", which has the translation $(exists x)[(forall y(L(x,y))),land,L(x,J)]$. Here $L(a,b)$ means $a$ likes $b$.
(The first is probably the intended meaning, because if somebody liked everyone, they would surely like John. I'm just pointing out an interpretation closer to what you were originally thinking.)
answered Apr 27 '17 at 2:52
shardulc
3,620927
edited Apr 29 '17 at 5:56
answered Apr 27 '17 at 2:52
shardulc
3,620927
answered Apr 27 '17 at 2:52
shardulc
3,620927
answered Apr 27 '17 at 2:52
shardulc
3,620927
3,620927
Thank you. The second answer is what I originally thinking. I was not sure how to translate it correctly. I agree with you that the first translation might be the intended meaning.
– Haley_Q
Apr 27 '17 at 3:14
@Haley_Q You're welcome! If you like my answer, please upvote it (the up arrow) and accept it if it answers your question.
– shardulc
Apr 27 '17 at 3:31
For the first interpretation, do you really want $y$ to be a free variable, or did you miss out a $forall y$?
– bof
Apr 29 '17 at 5:17
@bof Thanks, fixed it.
– shardulc
Apr 29 '17 at 5:56
add a comment |Â
Thank you. The second answer is what I originally thinking. I was not sure how to translate it correctly. I agree with you that the first translation might be the intended meaning.
– Haley_Q
Apr 27 '17 at 3:14
@Haley_Q You're welcome! If you like my answer, please upvote it (the up arrow) and accept it if it answers your question.
– shardulc
Apr 27 '17 at 3:31
For the first interpretation, do you really want $y$ to be a free variable, or did you miss out a $forall y$?
– bof
Apr 29 '17 at 5:17
@bof Thanks, fixed it.
– shardulc
Apr 29 '17 at 5:56
Thank you. The second answer is what I originally thinking. I was not sure how to translate it correctly. I agree with you that the first translation might be the intended meaning.
– Haley_Q
Apr 27 '17 at 3:14
Thank you. The second answer is what I originally thinking. I was not sure how to translate it correctly. I agree with you that the first translation might be the intended meaning.
– Haley_Q
Apr 27 '17 at 3:14
@Haley_Q You're welcome! If you like my answer, please upvote it (the up arrow) and accept it if it answers your question.
– shardulc
Apr 27 '17 at 3:31
@Haley_Q You're welcome! If you like my answer, please upvote it (the up arrow) and accept it if it answers your question.
– shardulc
Apr 27 '17 at 3:31
For the first interpretation, do you really want $y$ to be a free variable, or did you miss out a $forall y$?
– bof
Apr 29 '17 at 5:17
For the first interpretation, do you really want $y$ to be a free variable, or did you miss out a $forall y$?
– bof
Apr 29 '17 at 5:17
@bof Thanks, fixed it.
– shardulc
Apr 29 '17 at 5:56
@bof Thanks, fixed it.
– shardulc
Apr 29 '17 at 5:56
add a comment |Â
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I'm assuming $L(x,y)$ means $x$ likes $y$. Reading your thought to myself, I read: "There is somebody ($x$) who likes $y$ and who likes $J$."
– Mr Toad
Apr 27 '17 at 2:07