# Thread: help me to correct my exercise on nested quantifier

1. ## help me to correct my exercise on nested quantifier

Let F(x,y) be the statement "x can fool y", where the domain consists of all people in the world. Use quantifiers to express each of these statements.
a. Nancy can fool exactly two people

$\exists x\exists y\forall z [ (F(Nancy, x) \wedge F(Nancy, y)) \wedge ((z = x) \wedge (z = y))]$

b. No one can fool himself or herself

$\exists x\forall y [\neg F(x,x) \wedge ((y=x))]$

c. There is exactly one person whom everybody can fool

$\exists x\forall z [F(z,x) \wedge (x \neq z)]$

2. Originally Posted by TheRekz
Let F(x,y) be the statement "x can fool y", where the domain consists of all people in the world. Use quantifiers to express each of these statements.
a. Nancy can fool exactly two people
b. No one can fool himself or herself
c. There is exactly one person whom everybody can fool
I would say that we need more for part (a).
$\left( {\exists x} \right)\left( {\exists y} \right)\left( {\forall z} \right)\left[ {\left( {F(N,x) \wedge F(N,y) \wedge (x \not= y)} \right) \wedge \left( {F(N,z) \Rightarrow \left( {z = y \vee z = x} \right)} \right)} \right]$.

I think that you need less for part (b).
$\forall x [\neg F(x,x)]$

Part (c) is a bit more complicated.
$\left( {\exists x} \right)\left( {\forall y} \right)\left[ {F(y,x) \wedge \left[ {\left( {\exists z} \right)\left( {F(y,z)} \right) \Rightarrow \left( {x = z} \right)} \right]} \right]$.

Here is a disclaimer: No two instructors/textbooks will agree in all cases.

3. how about if the question is "No one can fool both Fred and Jerry"

$\forall x [\neg F(x,Fred) \wedge \neg F(x,Jerry)]$

and

Everyone can be fooled by somebody is:

$\forall x\exists y [F(y,x)]$

4. Originally Posted by TheRekz
how about if the question is "No one can fool both Fred and Jerry"

$\forall x \neg[F(x,Fred) \wedge F(x,Jerry)]$

and

Everyone can be fooled by somebody is:

$\forall x\exists y [F(y,x)]$
Yes, I would accept both of those.
Note that the first one is equivalent to:
$\neg \left( {\exists x} \right)\left[ {F(x,\text {Fred}) \wedge F(x,\text {Jerry)}} \right]$

5. Okay I have one more 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.

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

$\exists x\forall y [I(x) \wedge \neg C(x,y) \wedge x \neq y ]$

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

$\exists x\exists y\forall z [\neg C(x,y) \Rightarrow (x \neq y \wedge \ x \neq z \wedge \ y \neq z)]$

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

$\exists x \forall y [C(x,y) \wedge (y = x)]$

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

$\exists x \forall y [\neg I(x) \wedge I(y) \wedge (y \neq x)]$

6. a)yeah, it makes sense, the x net equal to is interesting since one might say that that is assumed( people dont usually talk to themselves), but its not wrong

b) why do you use z? if z represents all the students in the class saying $x \neq$ z or $y \neq z$ implies x and y are not in the class

7. c) youre going overboard with the implications here, what youre saying here is that if there exists a student x that has chatted with all students, than all the students are student x...

d) do you know of the unique existential quantifier, because it would come in handy here, if you dont know it, look for it in your book because it wil be defined usually in terms of the other two quantifers which is essentially what you want to do for this problem

8. Originally Posted by Ilaggoodly
a)yeah, it makes sense, the x net equal to is interesting since one might say that that is assumed( people dont usually talk to themselves), but its not wrong

b) why do you use z? if z represents all the students in the class saying $x \neq$ z or $y \neq z$ implies x and y are not in the class
so as for b I don't have to use the variable z? just erase it?

9. Originally Posted by Ilaggoodly
c) youre going overboard with the implications here, what youre saying here is that if there exists a student x that has chatted with all students, than all the students are student x...

d) do you know of the unique existential quantifier, because it would come in handy here, if you dont know it, look for it in your book because it wil be defined usually in terms of the other two quantifers which is essentially what you want to do for this problem

for part c. I have to erase y = x??

10. i would say yes to both