Question

Let I(x) be the statement “x has an Internet connection” and C(x, y) be the statement “x and y have chatted over the Internet,” where the domain for the variables x and y consists of all students in your class. Use quantifiers to express each of these statements.

a) Jerry does not have an Internet connection.

b) Rachel has not chatted over the Internet with Chelsea.

c) Jan and Sharon have never chatted over the Internet.

d) No one in the class has chatted with Bob.

e) Sanjay has chatted with everyone except Joseph.

f) Someone in your class does not have an Internet connection.

g) Not everyone in your class has an Internet connection.

h) Exactly one student in your class has an Internet connection.

i) Everyone except one student in your class has an Internet connection.

j) Everyone in your class with an Internet connection has chatted over the Internet with at least one other student in your class.

k) Someone in your class has an Internet connection but has not chatted with anyone else in your class.

l) There are two students in your class who have not chatted with each other over the Internet.

m) There is a student in your class who has chatted with everyone in your class over the Internet.

n) There are at least two students in your class who have not chatted with the same person in your class.

o) There are two students in the class who between them have chatted with everyone else in the class.

Short answer

a) $\mathbf{\neg }\mathbf{I}\mathbf{\left(}\mathbf{Jerry}\mathbf{\right)}$

b)$\mathbf{\neg }\mathbf{C}\mathbf{(}\mathbf{Rachel}\mathbf{,}\mathbf{Chelsea}\mathbf{)}$

c) $\mathbf{\neg }\mathbf{C}\mathbf{(}\mathbf{Jan}\mathbf{,}\mathbf{Sharon}\mathbf{)}$

d) $\mathbf{\forall }\mathbf{x}\mathbf{\neg }\mathbf{C}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{Bob}\mathbf{)}$

e) $\mathbf{\forall }\mathbf{y}\mathbf{(}\mathbf{y}\mathbf{\ne }\mathbf{Joseph}\mathbf{\to }\mathbf{C}\mathbf{(}\mathbf{Sanjay}\mathbf{,}\mathbf{y}\mathbf{\left)}\mathbf{\right)}$

f) $\mathbf{\exists }\mathbf{x}\mathbf{\neg }\mathbf{I}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

g) $\mathbf{\neg }\mathbf{\forall }\mathbf{xI}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

h) $\mathbf{\exists }\mathbf{x}\mathbf{\left(}\mathbf{I}\mathbf{\right(}\mathbf{x}\mathbf{)}\mathbf{\wedge }\mathbf{\forall }\mathbf{y}\mathbf{(}\mathbf{y}\mathbf{\ne }\mathbf{x}\mathbf{\to }\mathbf{\neg }\mathbf{I}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{\left)}\mathbf{\right)}$

i) $\mathbf{\exists }\mathbf{x}\mathbf{(}\mathbf{\neg }\mathbf{I}\mathbf{(}\mathbf{x}\mathbf{)}\mathbf{\wedge }\mathbf{\forall }\mathbf{y}\mathbf{(}\mathbf{y}\mathbf{\ne }\mathbf{x}\mathbf{\to }\mathbf{I}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{\left)}\mathbf{\right)}$

j)$\mathbf{\exists }\mathbf{x}\mathbf{\left(}\mathbf{I}\mathbf{\right(}\mathbf{x}\mathbf{)}\mathbf{\to }\mathbf{\exists }\mathbf{y}\mathbf{(}\mathbf{C}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{\wedge }\mathbf{x}\mathbf{\ne }\mathbf{y}\mathbf{\left)}\mathbf{\right)}$

k) $\mathbf{\exists }\mathbf{x}\mathbf{\left(}\mathbf{I}\mathbf{\right(}\mathbf{x}\mathbf{)}\mathbf{\wedge }\mathbf{\forall }\mathbf{y}\mathbf{\neg }\mathbf{C}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\left)}\mathbf{\right)}$

l) $\mathbf{\exists }\mathbf{x}\mathbf{\exists }\mathbf{y}\mathbf{(}\mathbf{x}\mathbf{\ne }\mathbf{y}\mathbf{\wedge }\mathbf{\neg }\mathbf{C}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\left)}\mathbf{\right)}$

m) $\mathbf{\exists }\mathbf{x}\mathbf{\forall }\mathbf{yC}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}$

n) $\mathbf{\exists }\mathbf{x}\mathbf{\exists }\mathbf{y}\mathbf{(}\mathbf{x}\mathbf{\ne }\mathbf{y}\mathbf{\wedge }\mathbf{\exists }\mathbf{z}\mathbf{(}\mathbf{x}\mathbf{\ne }\mathbf{z}\mathbf{\wedge }\mathbf{y}\mathbf{\ne }\mathbf{z}\mathbf{\wedge }\mathbf{\neg }\mathbf{C}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{\wedge }\mathbf{\neg }\mathbf{C}\mathbf{(}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{\left)}\mathbf{\right)}$

o)$\mathbf{\exists }\mathbf{x}\mathbf{\exists }\mathbf{y}\mathbf{(}\mathbf{x}\mathbf{\ne }\mathbf{y}\mathbf{\wedge }\mathbf{\forall }\mathbf{z}\mathbf{(}\mathbf{C}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{\vee }\mathbf{C}\mathbf{(}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{\left)}\mathbf{\right)}$

Step-by-step solution

Part (a)

$\mathbf{I}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$= “x has an internet connection”

Negation of the above can be given as:

$\mathbf{\neg }\mathbf{I}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$= “x does not have an internet connection”

So the given statement can be expressed as:

$\mathbf{\neg }\mathbf{I}\mathbf{\left(}\mathbf{Jerry}\mathbf{\right)}$

Part (b)

$\mathrm{C}(\mathrm{x},\mathrm{y})$= “x and y have chatted over the internet”

Negation of the above can be given as:

$\neg \mathrm{C}(\mathrm{x},\mathrm{y})$= “x has not chatted with y over the internet”

So the given statement can be expressed as:

$\neg \mathrm{C}(\mathrm{Rachel},\mathrm{Chelsea})$

Part (c)

$\mathrm{C}(\mathrm{x},\mathrm{y})$= “x and y have chatted over the internet”

Negation of the above can also be given as:

$\neg \mathrm{C}(\mathrm{x},\mathrm{y})$= “x has never chatted with y over the internet”

So the given statement can be expressed as:

$\neg \mathrm{C}(\mathrm{Jan},\mathrm{Sharon})$

Part (d)

$\mathrm{C}(\mathrm{x},\mathrm{Bob})$=“x and Bob have chatted over the internet”

Negation of the above can be given as:

$\neg \mathrm{C}(\mathrm{x},\mathrm{Bob})$= “x and Bob have not chatted over the internet”

Given that $\neg \mathrm{C}(\mathrm{x},\mathrm{Bob})$is true for all x, therefore, the given statement can be expressed as:

$\forall \mathrm{x}\neg \mathrm{C}(\mathrm{x},\mathrm{Bob})$

Part (e)

$\mathrm{C}(\mathrm{Sanjay},\mathrm{y})$=“Sanjay and y have chatted over the internet”

Given that for all y, width="85" height="21" role="math" style="max-width: none; vertical-align: -5px;" localid="1668604412391" $\mathrm{C}(\mathrm{Sanjay},\mathrm{y})$is true where $\mathrm{y}\ne \mathrm{Joseph}$. This can be expressed as:

$\forall \mathrm{y}(\mathrm{y}\ne \mathrm{Joseph}\to \mathrm{C}(\mathrm{Sanjay},\mathrm{y}\left)\right)$

Part (f)

$\mathrm{I}\left(\mathrm{x}\right)$= “x has an internet connection”

Given that there exists an x such that above statement is false. This can be expressed as:

$\exists \mathrm{x}\neg \mathrm{I}\left(\mathrm{x}\right)$

Part (g)

I(x)= “x has an internet connection”

Given thatis not true for all x. This can be expressed as:

$\neg \forall \mathrm{xI}\left(\mathrm{x}\right)$

Part (h)

$\mathrm{I}\left(\mathrm{x}\right)$= “x has an internet connection”

Given that there exist only one x such that $\mathrm{I}\left(\mathrm{x}\right)$is true, else it’s false. This can be expressed as:

$\exists \mathrm{x}\left(\mathrm{I}\right(\mathrm{x})\wedge \forall \mathrm{y}(\mathrm{y}\ne \mathrm{x}\to \neg \mathrm{I}\left(\mathrm{y}\right)\left)\right)$

Part (i)

I(x)= “x has an internet connection”

Given that there exists an x such that I(x) is false for exactly one x, else it’s true. This can be expressed as:

$\exists \mathrm{x}(\neg \mathrm{I}(\mathrm{x})\wedge \forall \mathrm{y}(\mathrm{y}\ne \mathrm{x}\to \mathrm{I}\left(\mathrm{y}\right)\left)\right)$

Part (j)

I(x)= “x has an internet connection”

C(x,y)= “x and y have chatted over the internet”

Given that there exists an x and a y (where $\mathrm{x}\ne \mathrm{y}$) such that if I(x) is true then C(x,y) is also true. This can be expressed as:

$\exists \mathrm{x}\left(\mathrm{I}\right(\mathrm{x})\to \exists \mathrm{y}(\mathrm{C}(\mathrm{x},\mathrm{y})\wedge \mathrm{x}\ne \mathrm{y}\left)\right)$

Part (k)

\(I(x)\)= “x has an internet connection”

\(C(x,y)\)= “x and y have chatted over the internet”

Given that there exists an x such that \(I(x)\)is true but \(C(x,y)\) is false for all y. This can be expressed as:\[\exists x(I(x) \wedge \forall y\neg C(x,y))\]

Part (l)

\(C(x,y)\)= “x and y have chatted over the internet”

Given that there exist x and y\((x \ne y)\)such that\(C(x,y)\)is false. This can be expressed as:

\(\exists x\exists y(x \ne y \wedge \neg C(x,y))\)

Part (m)

\(C(x,y)\)= “x and y have chatted over the internet”

Given that there exist an x such that\(C(x,y)\)is true for all y. This can be expressed as:

\(\exists x\forall yC(x,y)\)

Part (n)

\(\neg C(x,y)\)= “x and y have not chatted over the internet”

Given that there are 2 different students (let x and y) who have not chatted with same z over the internet, therefore, either\(\neg C(x,z)\)or\(\neg C(y,z)\)is true.

This can be represented as:

\(\exists x\exists y(x \ne y \wedge \exists z(x \ne z \wedge y \ne z \wedge \neg C(x,z) \wedge \neg C(y,z)))\)

Part (o)

\(C(x,y)\)= “x and y have chatted over the internet”

Given that there are 2 different students (let x and y) such that either of themhas chatted with all students over the internet, therefore, either\(C(x,z)\)or\(C(y,z)\)is true.

This can be represented as:

\(\exists x\exists y(x \ne y \wedge \forall z(C(x,z) \vee C(y,z)))\)