inkyvoyd
  • inkyvoyd
Logic one min inserting question
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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inkyvoyd
  • inkyvoyd
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"
inkyvoyd
  • inkyvoyd
I looked up the answers to the this problem, but I have no friggin clue how they got there.
Psymon
  • Psymon
If this is what I think it is, are we talking about things like "there exists a number x such that" blah blah?

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inkyvoyd
  • inkyvoyd
Statement: ∃x∃y( Q(x,y) ∧ ∀z ( z ≠ x ∨ ~Q(x,z))))
inkyvoyd
  • inkyvoyd
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
anonymous
  • anonymous
Did you come up with that statement?
inkyvoyd
  • inkyvoyd
No, and I only kind of understand why z is there.
anonymous
  • anonymous
The z is there to make sure there isn't another person with whom they spoke.
anonymous
  • anonymous
z would be hypothetically the other person, so it's saying such a person does not exist.
inkyvoyd
  • inkyvoyd
namely, when we say one other student we mean that Q(x,x) can be true or false?
inkyvoyd
  • inkyvoyd
but x=y is a possiblity
inkyvoyd
  • inkyvoyd
in which case x has both chatted with x as well as with y, meaning x has chatted with 2 people
inkyvoyd
  • inkyvoyd
or is Q(x,x)=F for all cases?
anonymous
  • anonymous
Technically this statement allows yourself to be the person you're talking with. It doesn't not say \(x\neq y\)
inkyvoyd
  • inkyvoyd
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
anonymous
  • anonymous
You just have a lot of freedom.
inkyvoyd
  • inkyvoyd
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
anonymous
  • anonymous
Negate it logically first.
inkyvoyd
  • inkyvoyd
meaning that if you have chatted with yourself then you have not chatted with anyone else?
anonymous
  • anonymous
Well that is a possibility.
anonymous
  • anonymous
But you don't necessarily have to note that out. It's a logical consequence of the statement.
inkyvoyd
  • inkyvoyd
because if you have chatted with yourself that's exactly one person you've chatted with
anonymous
  • anonymous
It could be that \(Q(x.x)\) always results in false because it has a \(x\neq y\) baked into it.
inkyvoyd
  • inkyvoyd
can you show me how to construct the statement from scratch again?
anonymous
  • anonymous
In other works, it could be that \(Q(x,y)\) means "\(x\) has talked to \(y\) and \(x\) is not \(y\)"
anonymous
  • anonymous
If you've already been given a statement, stick with the statement you were given. Do you have to make up the statement yourself?
inkyvoyd
  • inkyvoyd
that statement I have to make up
anonymous
  • anonymous
What is the statement do you have to make up?
inkyvoyd
  • inkyvoyd
"∃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
anonymous
  • anonymous
So that is the answer you found online?
inkyvoyd
  • inkyvoyd
yeah
inkyvoyd
  • inkyvoyd
uhm
inkyvoyd
  • inkyvoyd
sorry it's ∃x∃y( Q(x,y) ∧ ∀z ( z = x ∨ ~Q(x,z))))
inkyvoyd
  • inkyvoyd
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
inkyvoyd
  • inkyvoyd
so the original statement I was given was ∃x∃y( Q(x,y) ∧ ∀z ( z ≠ x → ~Q(x,z))))
inkyvoyd
  • inkyvoyd
oh wait http://math.stackexchange.com/questions/73300/negation-of-uniqueness-quantifier
anonymous
  • anonymous
Yeah, you want to build around that.
anonymous
  • anonymous
Use that to make \(y\) unique.
anonymous
  • anonymous
Then around that put ∃x ( [...] ∧ x ≠ y) where [...] is the uniqueness for y.

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