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All students in a class are friends with some students in all engineering faculty
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All students in a class are friends with some students in all engineering faculty.
My solution was Domain $x,y$ is a student Domain $z$ is engineering faculty $M(x)$ students in this class $S(x,y)$; $x$ is friends with $y$; $F(y,z)$ denotes the sentence "$y$ is from $z$."
$$forall x exists y forall z (m(x)longrightarrow s(x,y) land f(y,z)).$$
What I’m having trouble with is translating to English after negating:
I’m having trouble with is translating to English after negating:
∃x ∀y ∃z (m(x)^(¬s(x,y)∨¬f(y,z)))
Is it right if I say some students in this class are not friends with all students in some other faculty
discrete-mathematics quantifiers logic-translation
asked Jul 21 at 12:30
Shayma Faiz
83
add a comment |Â
up vote
1
down vote
favorite
All students in a class are friends with some students in all engineering faculty.
My solution was Domain $x,y$ is a student Domain $z$ is engineering faculty $M(x)$ students in this class $S(x,y)$; $x$ is friends with $y$; $F(y,z)$ denotes the sentence "$y$ is from $z$."
$$forall x exists y forall z (m(x)longrightarrow s(x,y) land f(y,z)).$$
What I’m having trouble with is translating to English after negating:
I’m having trouble with is translating to English after negating:
∃x ∀y ∃z (m(x)^(¬s(x,y)∨¬f(y,z)))
Is it right if I say some students in this class are not friends with all students in some other faculty
discrete-mathematics quantifiers logic-translation
asked Jul 21 at 12:30
Shayma Faiz
83
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
All students in a class are friends with some students in all engineering faculty.
My solution was Domain $x,y$ is a student Domain $z$ is engineering faculty $M(x)$ students in this class $S(x,y)$; $x$ is friends with $y$; $F(y,z)$ denotes the sentence "$y$ is from $z$."
$$forall x exists y forall z (m(x)longrightarrow s(x,y) land f(y,z)).$$
What I’m having trouble with is translating to English after negating:
I’m having trouble with is translating to English after negating:
∃x ∀y ∃z (m(x)^(¬s(x,y)∨¬f(y,z)))
Is it right if I say some students in this class are not friends with all students in some other faculty
discrete-mathematics quantifiers logic-translation
asked Jul 21 at 12:30
Shayma Faiz
83
All students in a class are friends with some students in all engineering faculty.
My solution was Domain $x,y$ is a student Domain $z$ is engineering faculty $M(x)$ students in this class $S(x,y)$; $x$ is friends with $y$; $F(y,z)$ denotes the sentence "$y$ is from $z$."
$$forall x exists y forall z (m(x)longrightarrow s(x,y) land f(y,z)).$$
What I’m having trouble with is translating to English after negating:
I’m having trouble with is translating to English after negating:
∃x ∀y ∃z (m(x)^(¬s(x,y)∨¬f(y,z)))
Is it right if I say some students in this class are not friends with all students in some other faculty
discrete-mathematics quantifiers logic-translation
asked Jul 21 at 12:30
Shayma Faiz
83
edited Jul 22 at 3:41
asked Jul 21 at 12:30
Shayma Faiz
83
asked Jul 21 at 12:30
Shayma Faiz
83
asked Jul 21 at 12:30
Shayma Faiz
83
83
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
We want to say something about all students in this class:
So, $forall x (M(x) to ...$
That is, for all students $x$, if $x$ is a student in this class, then ...
... then there exist(s) some student(s) $y$, such that for all engineering faculty $z,;F(y, z)$ and $x$ is friends with $y$.
$$forall xBig(M(x) to exists y forall zbig(S(x, y) land F(y, z)big)Big)tagtranslation$$
Your domains are fine. We use the predicate $M(x)$ to denote the student $x$ is in “this classâ€Â, and the predicate $F(y, z)$ to denote $y$ is from the engineering faculty $z$. Note that in the translation, the outermost quantifier quantifies $x$.
Negating the above, we have
$$lnot forall xBig(M(x) to exists y forall zbig(S(x, y)land F(y, z)big)Big)tag$lnot$ translation$$
$$equiv exists x lnotBig(lnot M(x) lor exists y forall zbig(S(x, y)land F(y, z)big)Big)tag$pto q equiv lnot p lor q$$$
$$exists xBig(M(x) land lnotbig(exists y forall z( S(x, y)land F(y, z)big)Big)tagDeMorgan's$$
$$equiv exists xBig( M(x) land forall y exists zbig(lnot S(x,y) lor lnot F(y, z)big)Big)tagDeMorgan's$$
$$equiv exists xBig(M(x) land forall y big( S(x, y) to exists z(lnot F(y, z))big)Big)tag$lnot p lor lnot q equiv p to lnot q$$$
This basically states: There are some students $x$ in this class and all of their friends are not from one or more of the engineering faculty.
edited Jul 23 at 11:13
José Carlos Santos
114k1698177
answered Jul 21 at 13:09
amWhy
189k25219431
Why did you put F(y,z) before S(x,y) i dont get that part, thank you so much for the answer I have finals tomorrow
– Shayma Faiz
Jul 22 at 3:21
Remember that conjunction is commutative. $pland q equiv qland p$. I just happened, at the start, to list F(y, z) before S(x, y). The other order would have been fine. Notice that in the end, I reversed the order using commutativity.
– amWhy
Jul 22 at 10:39
Shayma, I've started with the order of your translation, since that simplifies the rendering of the negated statement.
– amWhy
Jul 22 at 11:46
It really helped me alot ! I appreciate it thank you!! ☺ï¸Â😆
– Shayma Faiz
Jul 22 at 11:48
Glad to have helped!
– amWhy
Jul 22 at 11:51
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
accepted
We want to say something about all students in this class:
So, $forall x (M(x) to ...$
That is, for all students $x$, if $x$ is a student in this class, then ...
... then there exist(s) some student(s) $y$, such that for all engineering faculty $z,;F(y, z)$ and $x$ is friends with $y$.
$$forall xBig(M(x) to exists y forall zbig(S(x, y) land F(y, z)big)Big)tagtranslation$$
Your domains are fine. We use the predicate $M(x)$ to denote the student $x$ is in “this classâ€Â, and the predicate $F(y, z)$ to denote $y$ is from the engineering faculty $z$. Note that in the translation, the outermost quantifier quantifies $x$.
Negating the above, we have
$$lnot forall xBig(M(x) to exists y forall zbig(S(x, y)land F(y, z)big)Big)tag$lnot$ translation$$
$$equiv exists x lnotBig(lnot M(x) lor exists y forall zbig(S(x, y)land F(y, z)big)Big)tag$pto q equiv lnot p lor q$$$
$$exists xBig(M(x) land lnotbig(exists y forall z( S(x, y)land F(y, z)big)Big)tagDeMorgan's$$
$$equiv exists xBig( M(x) land forall y exists zbig(lnot S(x,y) lor lnot F(y, z)big)Big)tagDeMorgan's$$
$$equiv exists xBig(M(x) land forall y big( S(x, y) to exists z(lnot F(y, z))big)Big)tag$lnot p lor lnot q equiv p to lnot q$$$
This basically states: There are some students $x$ in this class and all of their friends are not from one or more of the engineering faculty.
edited Jul 23 at 11:13
José Carlos Santos
114k1698177
answered Jul 21 at 13:09
amWhy
189k25219431
Why did you put F(y,z) before S(x,y) i dont get that part, thank you so much for the answer I have finals tomorrow
– Shayma Faiz
Jul 22 at 3:21
Remember that conjunction is commutative. $pland q equiv qland p$. I just happened, at the start, to list F(y, z) before S(x, y). The other order would have been fine. Notice that in the end, I reversed the order using commutativity.
– amWhy
Jul 22 at 10:39
Shayma, I've started with the order of your translation, since that simplifies the rendering of the negated statement.
– amWhy
Jul 22 at 11:46
It really helped me alot ! I appreciate it thank you!! ☺ï¸Â😆
– Shayma Faiz
Jul 22 at 11:48
Glad to have helped!
– amWhy
Jul 22 at 11:51
add a comment |Â
up vote
1
down vote
accepted
We want to say something about all students in this class:
So, $forall x (M(x) to ...$
That is, for all students $x$, if $x$ is a student in this class, then ...
... then there exist(s) some student(s) $y$, such that for all engineering faculty $z,;F(y, z)$ and $x$ is friends with $y$.
$$forall xBig(M(x) to exists y forall zbig(S(x, y) land F(y, z)big)Big)tagtranslation$$
Your domains are fine. We use the predicate $M(x)$ to denote the student $x$ is in “this classâ€Â, and the predicate $F(y, z)$ to denote $y$ is from the engineering faculty $z$. Note that in the translation, the outermost quantifier quantifies $x$.
Negating the above, we have
$$lnot forall xBig(M(x) to exists y forall zbig(S(x, y)land F(y, z)big)Big)tag$lnot$ translation$$
$$equiv exists x lnotBig(lnot M(x) lor exists y forall zbig(S(x, y)land F(y, z)big)Big)tag$pto q equiv lnot p lor q$$$
$$exists xBig(M(x) land lnotbig(exists y forall z( S(x, y)land F(y, z)big)Big)tagDeMorgan's$$
$$equiv exists xBig( M(x) land forall y exists zbig(lnot S(x,y) lor lnot F(y, z)big)Big)tagDeMorgan's$$
$$equiv exists xBig(M(x) land forall y big( S(x, y) to exists z(lnot F(y, z))big)Big)tag$lnot p lor lnot q equiv p to lnot q$$$
This basically states: There are some students $x$ in this class and all of their friends are not from one or more of the engineering faculty.
edited Jul 23 at 11:13
José Carlos Santos
114k1698177
answered Jul 21 at 13:09
amWhy
189k25219431
Why did you put F(y,z) before S(x,y) i dont get that part, thank you so much for the answer I have finals tomorrow
– Shayma Faiz
Jul 22 at 3:21
Remember that conjunction is commutative. $pland q equiv qland p$. I just happened, at the start, to list F(y, z) before S(x, y). The other order would have been fine. Notice that in the end, I reversed the order using commutativity.
– amWhy
Jul 22 at 10:39
Shayma, I've started with the order of your translation, since that simplifies the rendering of the negated statement.
– amWhy
Jul 22 at 11:46
It really helped me alot ! I appreciate it thank you!! ☺ï¸Â😆
– Shayma Faiz
Jul 22 at 11:48
Glad to have helped!
– amWhy
Jul 22 at 11:51
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We want to say something about all students in this class:
So, $forall x (M(x) to ...$
That is, for all students $x$, if $x$ is a student in this class, then ...
... then there exist(s) some student(s) $y$, such that for all engineering faculty $z,;F(y, z)$ and $x$ is friends with $y$.
$$forall xBig(M(x) to exists y forall zbig(S(x, y) land F(y, z)big)Big)tagtranslation$$
Your domains are fine. We use the predicate $M(x)$ to denote the student $x$ is in “this classâ€Â, and the predicate $F(y, z)$ to denote $y$ is from the engineering faculty $z$. Note that in the translation, the outermost quantifier quantifies $x$.
Negating the above, we have
$$lnot forall xBig(M(x) to exists y forall zbig(S(x, y)land F(y, z)big)Big)tag$lnot$ translation$$
$$equiv exists x lnotBig(lnot M(x) lor exists y forall zbig(S(x, y)land F(y, z)big)Big)tag$pto q equiv lnot p lor q$$$
$$exists xBig(M(x) land lnotbig(exists y forall z( S(x, y)land F(y, z)big)Big)tagDeMorgan's$$
$$equiv exists xBig( M(x) land forall y exists zbig(lnot S(x,y) lor lnot F(y, z)big)Big)tagDeMorgan's$$
$$equiv exists xBig(M(x) land forall y big( S(x, y) to exists z(lnot F(y, z))big)Big)tag$lnot p lor lnot q equiv p to lnot q$$$
This basically states: There are some students $x$ in this class and all of their friends are not from one or more of the engineering faculty.
edited Jul 23 at 11:13
José Carlos Santos
114k1698177
answered Jul 21 at 13:09
amWhy
189k25219431
We want to say something about all students in this class:
So, $forall x (M(x) to ...$
That is, for all students $x$, if $x$ is a student in this class, then ...
... then there exist(s) some student(s) $y$, such that for all engineering faculty $z,;F(y, z)$ and $x$ is friends with $y$.
$$forall xBig(M(x) to exists y forall zbig(S(x, y) land F(y, z)big)Big)tagtranslation$$
Your domains are fine. We use the predicate $M(x)$ to denote the student $x$ is in “this classâ€Â, and the predicate $F(y, z)$ to denote $y$ is from the engineering faculty $z$. Note that in the translation, the outermost quantifier quantifies $x$.
Negating the above, we have
$$lnot forall xBig(M(x) to exists y forall zbig(S(x, y)land F(y, z)big)Big)tag$lnot$ translation$$
$$equiv exists x lnotBig(lnot M(x) lor exists y forall zbig(S(x, y)land F(y, z)big)Big)tag$pto q equiv lnot p lor q$$$
$$exists xBig(M(x) land lnotbig(exists y forall z( S(x, y)land F(y, z)big)Big)tagDeMorgan's$$
$$equiv exists xBig( M(x) land forall y exists zbig(lnot S(x,y) lor lnot F(y, z)big)Big)tagDeMorgan's$$
$$equiv exists xBig(M(x) land forall y big( S(x, y) to exists z(lnot F(y, z))big)Big)tag$lnot p lor lnot q equiv p to lnot q$$$
This basically states: There are some students $x$ in this class and all of their friends are not from one or more of the engineering faculty.
edited Jul 23 at 11:13
José Carlos Santos
114k1698177
answered Jul 21 at 13:09
amWhy
189k25219431
edited Jul 23 at 11:13
José Carlos Santos
114k1698177
edited Jul 23 at 11:13
José Carlos Santos
114k1698177
edited Jul 23 at 11:13
José Carlos Santos
114k1698177
114k1698177
answered Jul 21 at 13:09
amWhy
189k25219431
answered Jul 21 at 13:09
amWhy
189k25219431
answered Jul 21 at 13:09
amWhy
189k25219431
189k25219431
Why did you put F(y,z) before S(x,y) i dont get that part, thank you so much for the answer I have finals tomorrow
– Shayma Faiz
Jul 22 at 3:21
Remember that conjunction is commutative. $pland q equiv qland p$. I just happened, at the start, to list F(y, z) before S(x, y). The other order would have been fine. Notice that in the end, I reversed the order using commutativity.
– amWhy
Jul 22 at 10:39
Shayma, I've started with the order of your translation, since that simplifies the rendering of the negated statement.
– amWhy
Jul 22 at 11:46
It really helped me alot ! I appreciate it thank you!! ☺ï¸Â😆
– Shayma Faiz
Jul 22 at 11:48
Glad to have helped!
– amWhy
Jul 22 at 11:51
add a comment |Â
Why did you put F(y,z) before S(x,y) i dont get that part, thank you so much for the answer I have finals tomorrow
– Shayma Faiz
Jul 22 at 3:21
Remember that conjunction is commutative. $pland q equiv qland p$. I just happened, at the start, to list F(y, z) before S(x, y). The other order would have been fine. Notice that in the end, I reversed the order using commutativity.
– amWhy
Jul 22 at 10:39
Shayma, I've started with the order of your translation, since that simplifies the rendering of the negated statement.
– amWhy
Jul 22 at 11:46
It really helped me alot ! I appreciate it thank you!! ☺ï¸Â😆
– Shayma Faiz
Jul 22 at 11:48
Glad to have helped!
– amWhy
Jul 22 at 11:51
Why did you put F(y,z) before S(x,y) i dont get that part, thank you so much for the answer I have finals tomorrow
– Shayma Faiz
Jul 22 at 3:21
Why did you put F(y,z) before S(x,y) i dont get that part, thank you so much for the answer I have finals tomorrow
– Shayma Faiz
Jul 22 at 3:21
Remember that conjunction is commutative. $pland q equiv qland p$. I just happened, at the start, to list F(y, z) before S(x, y). The other order would have been fine. Notice that in the end, I reversed the order using commutativity.
– amWhy
Jul 22 at 10:39
Remember that conjunction is commutative. $pland q equiv qland p$. I just happened, at the start, to list F(y, z) before S(x, y). The other order would have been fine. Notice that in the end, I reversed the order using commutativity.
– amWhy
Jul 22 at 10:39
Shayma, I've started with the order of your translation, since that simplifies the rendering of the negated statement.
– amWhy
Jul 22 at 11:46
Shayma, I've started with the order of your translation, since that simplifies the rendering of the negated statement.
– amWhy
Jul 22 at 11:46
It really helped me alot ! I appreciate it thank you!! ☺ï¸Â😆
– Shayma Faiz
Jul 22 at 11:48
It really helped me alot ! I appreciate it thank you!! ☺ï¸Â😆
– Shayma Faiz
Jul 22 at 11:48
Glad to have helped!
– amWhy
Jul 22 at 11:51
Glad to have helped!
– amWhy
Jul 22 at 11:51
add a comment |Â
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