inkyvoyd
  • inkyvoyd
Discrete mathematics (usage of the existential and universal quantifiers and translation into the english language. I'll medal and fan an accurate answerer, whatever wings your ding
Mathematics
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jamiebookeater
  • jamiebookeater
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inkyvoyd
  • inkyvoyd
Hello, can someone please check my work for the following question? "Let Q(x,y) be the statement "student x has been a contestant on quiz show y." Express each of these sentences in terms of Q(x, y), quantifiers, and logical connectives, where the universe of discourse for x consists of all students at your school and for y consists of all quiz shows on television.
inkyvoyd
  • inkyvoyd
b. "No student at your school has ever been a contestant on a television quiz show" I got ¬∃x∃yQ(x,y). Is this correct in terms of operator precedence?
inkyvoyd
  • inkyvoyd
@wio welp?

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anonymous
  • anonymous
I don't think \( \exists y \) should be used here.
inkyvoyd
  • inkyvoyd
I have to use a binding for y otherwise it remains a predicate though
anonymous
  • anonymous
Hmmm, yeah I suppose it is ok.
inkyvoyd
  • inkyvoyd
the problem is I don't want to say ~(∃x∃y)Q(x,y)
anonymous
  • anonymous
Yeah, that is the problem I originally saw is that ambiguity.
inkyvoyd
  • inkyvoyd
add my own parenthesis?
anonymous
  • anonymous
There is also \( \forall x\neg \exists yQ(x,y) \)
inkyvoyd
  • inkyvoyd
are you sure that's the same thing?
inkyvoyd
  • inkyvoyd
∀x¬∃yQ(x,y)=∀x∀y¬Q(x,y)
anonymous
  • anonymous
It's not as direct because it is saying "All students have not been on any quiz show."
anonymous
  • anonymous
It's logically equivalent. I suppose you could just try doing parenthesis.
inkyvoyd
  • inkyvoyd
wait google http://site.iugaza.edu.ps/amarasa/files/sec-1.4.pdf

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