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Let Q(x,y) be the statement “x has been a contestant on quiz show y”, where the domain of x is the set of students and the domain for y consists of all quiz shows. For each of the English sentences below, please express it in terms of $Q(x,y)$ with quantifiers.

Please Correct me:

(i) Alice has never appeared in Jeopardy. $\quad\exists x \exists y \neg Q(x,y) $

(ii) Every quiz show has had a student as a contestant. $\quad \forall y \exists x Q(x,y)$

(iii) No student has appeared in both Wheel of Fortune and Family Feud:

  • $\neg \exists x_1\neg \exists x_2 Q((x_1, \text{Wheel of Fortune}) \land \exists x_2 Q(x_2, \text{Family Feud}))$

2 Answers 2

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  • Let $Q(x,y)$ be the statement “$x$ has been a contestant on quiz show $y$”,

    • the domain of $x$ is the set of students and

    • the domain for $y$ consists of all quiz shows.


(i) Alice has never appeared in Jeopardy.

  • Let $a$: Alice (assuming Alice is a student).
  • Let $j$: jeopardy.

  • Then we have, $\lnot Q(a, j)\tag{i}$

  • Given (i), it is certainly true that your translation "$\exists x \exists y \neg Q(x,y)$" follows from (i): "There is some student x and some quiz show y such that $Q(x, y)$".

    • But your translation is not a translation of the given sentence: Your translation says nothing to the effect that "Alice has never been a contestant of jeopardy".
    • A quantifier is not appropriate* here. We need only the predicate $Q(x, y)$, where "x" is replaced by a named constant for "Alice", and "y" is replaced by a named constant for "Jeopardy".

(ii) Every quiz show has had a student as a contestant.

  • Yes, you're correct: $\forall y \exists x Q(x,y)\tag{ii}$Nice work!

(iii) No student has appeared in both Wheel of Fortune and Family Feud.

  • We need only one variable to represent a student, and we need only one quantifier:
    Here, we want to say something like:

    • "There does not exist a student x (or there is no student x) such that (x has appeared in Wheel of Fortune AND x has appeared in Family Feud)".
    • Equivalently, we can state "for all students x, it is not the case that (x has appeared in Wheel of Fortune AND x has appeared in Family Feud)."
  • To simplify matters, let's let

    • $f: $ Family feud
    • $w: $ Wheel of Fortune

$\neg\exists x (Q(x, w) \land Q(x, f))\quad\equiv\quad \forall x \lnot(Q(x, w) \land Q(x, f))\tag{iii}$

  • Can you see why your answer for (iii) is problematic? Try translating it into natural language and see if it matches the original sentence.
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    @PeterSmith (a) perhaps hopelessly suspicious? Personally, I try to give the benefit of doubt, but I've reached the "hopelessly suspicious" point, too. Perhaps I haven't taught long enough yet! Typically, if you follow my answers, I prefer to provide hints..., or ask leading questions. (b) Fair enough, yes you did give a couple of hints. (c) I do try to avoid words like "obviously" or "plainly" - but again, that's because I tend to trust that questions are genuine. (If it were "plainly" clear to the OP, why post?)2012-12-19
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Hints.

(i) is plainly wrong -- you need a constant denoting Jeopardy rather than the second quantifier.

(iii) is plainly wrong -- what you have written implies no one has ever appeared in Wheel of Fortune. You only need a single quantifier.

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    (i) Why leave out the need for a constant denoting Alice?2012-12-19