Question

Logic

one min inserting question

Answer 1

Express each of these statements using quantifiers. Then form the negation of the statement so that no negation is to the left of a quantifier. Next, express the negation in simple Eglish.
b. "There is a student in this class who has chatted with exactly one other student"

Answer 2

I looked up the answers to the this problem, but I have no friggin clue how they got there.

Answer 3

If this is what I think it is, are we talking about things like "there exists a number x such that" blah blah?

Answer 4

Statement: ∃x∃y( Q(x,y) ∧ ∀z ( z ≠ x ∨ ~Q(x,z))))

Answer 5

given the universe of discourse being defined as all students in this class, and Q(x,y) being defined as x has chatted with y

Answer 6

Did you come up with that statement?

Answer 7

No, and I only kind of understand why z is there.

Answer 8

The z is there to make sure there isn't another person with whom they spoke.

Answer 9

z would be hypothetically the other person, so it's saying such a person does not exist.

Answer 10

namely, when we say one other student we mean that Q(x,x) can be true or false?

Answer 11

but x=y is a possiblity

Answer 12

in which case x has both chatted with x as well as with y, meaning x has chatted with 2 people

Answer 13

or is Q(x,x)=F for all cases?

Answer 14

Technically this statement allows yourself to be the person you're talking with. It doesn't not say \(x\neq y\)

Answer 15

but the main issue is after I negate teh statement I have to translate it into english (which isn't a big deal gettign wrong since I have the answer key), but I have no clue how to do it

Answer 16

You just have a lot of freedom.

Answer 17

so you're saying that based off of the original english (the statement was not part of the problem, but I believe it is part of the standard answer key) you can be the person you have chatted with

Answer 18

Negate it logically first.

Answer 19

meaning that if you have chatted with yourself then you have not chatted with anyone else?

Answer 20

Well that is a possibility.

Answer 21

But you don't necessarily have to note that out. It's a logical consequence of the statement.

Answer 22

because if you have chatted with yourself that's exactly one person you've chatted with

Answer 23

It could be that \(Q(x.x)\) always results in false because it has a \(x\neq y\) baked into it.

Answer 24

can you show me how to construct the statement from scratch again?

Answer 25

In other works, it could be that \(Q(x,y)\) means "\(x\) has talked to \(y\) and \(x\) is not \(y\)"

Answer 26

If you've already been given a statement, stick with the statement you were given.

Do you have to make up the statement yourself?

Answer 27

that statement I have to make up

Answer 28

What is the statement do you have to make up?

Answer 29

"∃x∃y( Q(x,y) ∧ ∀z ( z ≠ x ∨ ~Q(x,z))))" that's the first part of the problem -I translate the engilsh sentence into a statement. I looked at a lot of logically equiavlent statements online by searching for the answers to the problem, and this one made the most sense to me

Answer 30

So that is the answer you found online?

Answer 31

yeah

Answer 32

uhm

Answer 33

sorry it's ∃x∃y( Q(x,y) ∧ ∀z ( z = x ∨ ~Q(x,z))))

Answer 34

the last disjunction was originally given in implication form p -> q but I changed it into ~p V q since it was just all a mess to me

Answer 35

so the original statement I was given was ∃x∃y( Q(x,y) ∧ ∀z ( z ≠ x → ~Q(x,z))))

Answer 36

oh wait
http://math.stackexchange.com/questions/73300/negation-of-uniqueness-quantifier

Answer 37

Yeah, you want to build around that.

Answer 38

Use that to make \(y\) unique.

Answer 39

Then around that put ∃x ( [...] ∧ x ≠ y) where [...] is the uniqueness for y.