Mixing
Sentences into LTL
Clash Royale CLAN TAG#URR8PPP
up vote
-1
down vote
favorite
Is this the correct conversion to LTL formula
Lions and zebras never drink at the waterhole together.
(Lions -> Drink) U (zebra -> Drink)
Even if you ask her infinitely often, she will never tell you.
GFask -> ¬Ftell
Once red, the light becomes green eventually after being yellow for a while. FRed -> Xyellow -> Fgreen
logic
asked 9 hours ago
Philip D'Souza
213
add a comment |Â
up vote
-1
down vote
favorite
Is this the correct conversion to LTL formula
Lions and zebras never drink at the waterhole together.
(Lions -> Drink) U (zebra -> Drink)
Even if you ask her infinitely often, she will never tell you.
GFask -> ¬Ftell
Once red, the light becomes green eventually after being yellow for a while. FRed -> Xyellow -> Fgreen
logic
asked 9 hours ago
Philip D'Souza
213
What is LTL? Do you have a reference?
– Bram28
9 hours ago
Linear Temporal Logic
– Philip D'Souza
9 hours ago
I downvoted, because of poor formatting and no context.
– Alex Kruckman
7 hours ago
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Is this the correct conversion to LTL formula
Lions and zebras never drink at the waterhole together.
(Lions -> Drink) U (zebra -> Drink)
Even if you ask her infinitely often, she will never tell you.
GFask -> ¬Ftell
Once red, the light becomes green eventually after being yellow for a while. FRed -> Xyellow -> Fgreen
logic
asked 9 hours ago
Philip D'Souza
213
Is this the correct conversion to LTL formula
Lions and zebras never drink at the waterhole together.
(Lions -> Drink) U (zebra -> Drink)
Even if you ask her infinitely often, she will never tell you.
GFask -> ¬Ftell
Once red, the light becomes green eventually after being yellow for a while. FRed -> Xyellow -> Fgreen
logic
asked 9 hours ago
Philip D'Souza
213
asked 9 hours ago
Philip D'Souza
213
asked 9 hours ago
Philip D'Souza
213
asked 9 hours ago
Philip D'Souza
213
213
What is LTL? Do you have a reference?
– Bram28
9 hours ago
Linear Temporal Logic
– Philip D'Souza
9 hours ago
I downvoted, because of poor formatting and no context.
– Alex Kruckman
7 hours ago
add a comment |Â
What is LTL? Do you have a reference?
– Bram28
9 hours ago
Linear Temporal Logic
– Philip D'Souza
9 hours ago
I downvoted, because of poor formatting and no context.
– Alex Kruckman
7 hours ago
What is LTL? Do you have a reference?
– Bram28
9 hours ago
What is LTL? Do you have a reference?
– Bram28
9 hours ago
Linear Temporal Logic
– Philip D'Souza
9 hours ago
Linear Temporal Logic
– Philip D'Souza
9 hours ago
I downvoted, because of poor formatting and no context.
– Alex Kruckman
7 hours ago
I downvoted, because of poor formatting and no context.
– Alex Kruckman
7 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
I have no experience with LTL, but I'm looking at the Wikipedia page, and do see some issues with your translations:
For your first one: Drink should be a predicate, of which lions and zebras (and other things) are the subject. So, use $Drink(lions)$ or $exists x (Lion(x) land Drink(x))$. Also, using $cup$ you end up saying that lions are drinking (at the waterhole) until zebras are drinking there ... but it is of course perfectly acceptable for none of them to drink there.
In fact, using $cup$ seems more appropriate for the third sentence, don't you think? Rather than using thew $rightarrow$ there.
And for the second one .. when it says 'even if you ask infinitely many times' ... I think it is just emphasizing the point that she never tells. Indeed, by using $G F ask -> ¬F tell$, you are saying that it is true that she never tells if she gets asked infinitely many times .. so that does not rule out that she does tell when not asked infitely many times, and I would say that the sentence means that she never tells, whether you ask here one, wtice, never, or infintiely many times. In other words, the whole bit about asking just seemsirrelevant, as it is not a condition of her not telling.
answered 9 hours ago
Bram28
54.5k33776
Aren't you assuming that none of them can drink there?
– Philip D'Souza
8 hours ago
Well, you need to symbolize that they can't drink together (or at least, that you can't have a lion and a zebra drinking at the same time). Saying that is not the same as saying that none of them can drink there at all. You can have a liion drink there, then a zerba, then two lions, etc tec ... as lonh as they don't drink together it's all ok . which is why I would not use the $cup$
– Bram28
8 hours ago
So something like (Drink(Lion)∧ ¬ Drink(Zebra)) U (Drink(Zebra)∧ ¬ Drink(lion))
– Philip D'Souza
8 hours ago
"Never" means "globally not". @PhilipD'Souza
– Graham Kemp
2 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I have no experience with LTL, but I'm looking at the Wikipedia page, and do see some issues with your translations:
For your first one: Drink should be a predicate, of which lions and zebras (and other things) are the subject. So, use $Drink(lions)$ or $exists x (Lion(x) land Drink(x))$. Also, using $cup$ you end up saying that lions are drinking (at the waterhole) until zebras are drinking there ... but it is of course perfectly acceptable for none of them to drink there.
In fact, using $cup$ seems more appropriate for the third sentence, don't you think? Rather than using thew $rightarrow$ there.
And for the second one .. when it says 'even if you ask infinitely many times' ... I think it is just emphasizing the point that she never tells. Indeed, by using $G F ask -> ¬F tell$, you are saying that it is true that she never tells if she gets asked infinitely many times .. so that does not rule out that she does tell when not asked infitely many times, and I would say that the sentence means that she never tells, whether you ask here one, wtice, never, or infintiely many times. In other words, the whole bit about asking just seemsirrelevant, as it is not a condition of her not telling.
answered 9 hours ago
Bram28
54.5k33776
Aren't you assuming that none of them can drink there?
– Philip D'Souza
8 hours ago
Well, you need to symbolize that they can't drink together (or at least, that you can't have a lion and a zebra drinking at the same time). Saying that is not the same as saying that none of them can drink there at all. You can have a liion drink there, then a zerba, then two lions, etc tec ... as lonh as they don't drink together it's all ok . which is why I would not use the $cup$
– Bram28
8 hours ago
So something like (Drink(Lion)∧ ¬ Drink(Zebra)) U (Drink(Zebra)∧ ¬ Drink(lion))
– Philip D'Souza
8 hours ago
"Never" means "globally not". @PhilipD'Souza
– Graham Kemp
2 hours ago
add a comment |Â
up vote
1
down vote
I have no experience with LTL, but I'm looking at the Wikipedia page, and do see some issues with your translations:
For your first one: Drink should be a predicate, of which lions and zebras (and other things) are the subject. So, use $Drink(lions)$ or $exists x (Lion(x) land Drink(x))$. Also, using $cup$ you end up saying that lions are drinking (at the waterhole) until zebras are drinking there ... but it is of course perfectly acceptable for none of them to drink there.
In fact, using $cup$ seems more appropriate for the third sentence, don't you think? Rather than using thew $rightarrow$ there.
And for the second one .. when it says 'even if you ask infinitely many times' ... I think it is just emphasizing the point that she never tells. Indeed, by using $G F ask -> ¬F tell$, you are saying that it is true that she never tells if she gets asked infinitely many times .. so that does not rule out that she does tell when not asked infitely many times, and I would say that the sentence means that she never tells, whether you ask here one, wtice, never, or infintiely many times. In other words, the whole bit about asking just seemsirrelevant, as it is not a condition of her not telling.
answered 9 hours ago
Bram28
54.5k33776
Aren't you assuming that none of them can drink there?
– Philip D'Souza
8 hours ago
Well, you need to symbolize that they can't drink together (or at least, that you can't have a lion and a zebra drinking at the same time). Saying that is not the same as saying that none of them can drink there at all. You can have a liion drink there, then a zerba, then two lions, etc tec ... as lonh as they don't drink together it's all ok . which is why I would not use the $cup$
– Bram28
8 hours ago
So something like (Drink(Lion)∧ ¬ Drink(Zebra)) U (Drink(Zebra)∧ ¬ Drink(lion))
– Philip D'Souza
8 hours ago
"Never" means "globally not". @PhilipD'Souza
– Graham Kemp
2 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I have no experience with LTL, but I'm looking at the Wikipedia page, and do see some issues with your translations:
For your first one: Drink should be a predicate, of which lions and zebras (and other things) are the subject. So, use $Drink(lions)$ or $exists x (Lion(x) land Drink(x))$. Also, using $cup$ you end up saying that lions are drinking (at the waterhole) until zebras are drinking there ... but it is of course perfectly acceptable for none of them to drink there.
In fact, using $cup$ seems more appropriate for the third sentence, don't you think? Rather than using thew $rightarrow$ there.
And for the second one .. when it says 'even if you ask infinitely many times' ... I think it is just emphasizing the point that she never tells. Indeed, by using $G F ask -> ¬F tell$, you are saying that it is true that she never tells if she gets asked infinitely many times .. so that does not rule out that she does tell when not asked infitely many times, and I would say that the sentence means that she never tells, whether you ask here one, wtice, never, or infintiely many times. In other words, the whole bit about asking just seemsirrelevant, as it is not a condition of her not telling.
answered 9 hours ago
Bram28
54.5k33776
I have no experience with LTL, but I'm looking at the Wikipedia page, and do see some issues with your translations:
For your first one: Drink should be a predicate, of which lions and zebras (and other things) are the subject. So, use $Drink(lions)$ or $exists x (Lion(x) land Drink(x))$. Also, using $cup$ you end up saying that lions are drinking (at the waterhole) until zebras are drinking there ... but it is of course perfectly acceptable for none of them to drink there.
In fact, using $cup$ seems more appropriate for the third sentence, don't you think? Rather than using thew $rightarrow$ there.
And for the second one .. when it says 'even if you ask infinitely many times' ... I think it is just emphasizing the point that she never tells. Indeed, by using $G F ask -> ¬F tell$, you are saying that it is true that she never tells if she gets asked infinitely many times .. so that does not rule out that she does tell when not asked infitely many times, and I would say that the sentence means that she never tells, whether you ask here one, wtice, never, or infintiely many times. In other words, the whole bit about asking just seemsirrelevant, as it is not a condition of her not telling.
answered 9 hours ago
Bram28
54.5k33776
edited 8 hours ago
answered 9 hours ago
Bram28
54.5k33776
answered 9 hours ago
Bram28
54.5k33776
answered 9 hours ago
Bram28
54.5k33776
54.5k33776
Aren't you assuming that none of them can drink there?
– Philip D'Souza
8 hours ago
Well, you need to symbolize that they can't drink together (or at least, that you can't have a lion and a zebra drinking at the same time). Saying that is not the same as saying that none of them can drink there at all. You can have a liion drink there, then a zerba, then two lions, etc tec ... as lonh as they don't drink together it's all ok . which is why I would not use the $cup$
– Bram28
8 hours ago
So something like (Drink(Lion)∧ ¬ Drink(Zebra)) U (Drink(Zebra)∧ ¬ Drink(lion))
– Philip D'Souza
8 hours ago
"Never" means "globally not". @PhilipD'Souza
– Graham Kemp
2 hours ago
add a comment |Â
Aren't you assuming that none of them can drink there?
– Philip D'Souza
8 hours ago
Well, you need to symbolize that they can't drink together (or at least, that you can't have a lion and a zebra drinking at the same time). Saying that is not the same as saying that none of them can drink there at all. You can have a liion drink there, then a zerba, then two lions, etc tec ... as lonh as they don't drink together it's all ok . which is why I would not use the $cup$
– Bram28
8 hours ago
So something like (Drink(Lion)∧ ¬ Drink(Zebra)) U (Drink(Zebra)∧ ¬ Drink(lion))
– Philip D'Souza
8 hours ago
"Never" means "globally not". @PhilipD'Souza
– Graham Kemp
2 hours ago
Aren't you assuming that none of them can drink there?
– Philip D'Souza
8 hours ago
Aren't you assuming that none of them can drink there?
– Philip D'Souza
8 hours ago
Well, you need to symbolize that they can't drink together (or at least, that you can't have a lion and a zebra drinking at the same time). Saying that is not the same as saying that none of them can drink there at all. You can have a liion drink there, then a zerba, then two lions, etc tec ... as lonh as they don't drink together it's all ok . which is why I would not use the $cup$
– Bram28
8 hours ago
Well, you need to symbolize that they can't drink together (or at least, that you can't have a lion and a zebra drinking at the same time). Saying that is not the same as saying that none of them can drink there at all. You can have a liion drink there, then a zerba, then two lions, etc tec ... as lonh as they don't drink together it's all ok . which is why I would not use the $cup$
– Bram28
8 hours ago
So something like (Drink(Lion)∧ ¬ Drink(Zebra)) U (Drink(Zebra)∧ ¬ Drink(lion))
– Philip D'Souza
8 hours ago
So something like (Drink(Lion)∧ ¬ Drink(Zebra)) U (Drink(Zebra)∧ ¬ Drink(lion))
– Philip D'Souza
8 hours ago
"Never" means "globally not". @PhilipD'Souza
– Graham Kemp
2 hours ago
"Never" means "globally not". @PhilipD'Souza
– Graham Kemp
2 hours ago
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873253%2fsentences-into-ltl%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
What is LTL? Do you have a reference?
– Bram28
9 hours ago
Linear Temporal Logic
– Philip D'Souza
9 hours ago
I downvoted, because of poor formatting and no context.
– Alex Kruckman
7 hours ago