# Express each of these sentences in terms of $Q(x, y)$, quantifiers, and logical connectives,

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.

(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:

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

• 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.
• Yes of course -- but here surely we should be giving hints for elementary homework, not saving the poster the effort of doing any thinking .... – Peter Smith Dec 19 '12 at 16:29
• I'd suggest for (ii): $\forall y\exists x\colon S(x)\land Q(x,y)$, where $S(x)$ denoets "$x$ is a student". – Hagen von Eitzen Dec 19 '12 at 16:29
• @Hagen The domain for x is "all students". – Namaste Dec 19 '12 at 16:38
• @Peter it appears to me that the asker has applied some effort to the problem. Pointing out what's wrong, in and of itself, does not necessarily help. – Namaste Dec 19 '12 at 16:39
• @amWhy (a) Call me an old cynic, but the "effort" seemed quite suspiciously hopeless, according to the plan "bung down anything and then ask for the right answer". (b) And I did give a hint or two. Yours, Old Cynic. – Peter Smith Dec 19 '12 at 19:44

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.

• (i) Why leave out the need for a constant denoting Alice? – Namaste Dec 19 '12 at 16:37