$ S(x) $ is the predicate "$x$ is a student"

$F(x)$ is the predicate "$x$ is a faculty member"

$A(x,y)$ is the predicate $x$ asked $y$ a question

I need to translate this sentence into logic: Some students asked every faculty member a question.

This is the answer: $\forall y(F(y) \longrightarrow \exists x(S(x) \vee A(x,y))) $

It seems to translate into: For all $y$, if $y$ is a faculty member, then there is some $x$ where $x$ is either a student or $x$ asks $y$ a question. This means that it's possible for $x$ to not be a student, but this makes no sense. I would replace $\vee$ with $\wedge $, would I be wrong?

My second question iswhether the following would also be correct, assuming the first answer is right:

$\exists x(S(x) \longrightarrow \forall y(F(y) \vee A(x,y)))$

There exists an $x$ where if $x$ is a student, then for all $y$ where $y$ is a faculty member, or (I still think it makes more sense if you plug and and here) $x$ asks $y$ a question. It seems to say the same thing.

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    $\begingroup$ Generally, you don't see expressions of the form $(\exists x)(* \to *)$; you tend to see $(\exists x)(* \wedge *)$. That's because if you had someone who wasn't a student, that person would be "proof" of your formula. ... Also, you tend to see $(\forall x)(* \to *)$. $\endgroup$ – Christopher Carl Heckman May 14 '16 at 21:52
  • $\begingroup$ I'm not sure that the book's answer is correct. It's really saying that every faculty member was asked a question, or for every faculty member, there is a student. And I've already discussed why your answer isn't right. $\endgroup$ – Christopher Carl Heckman May 14 '16 at 21:55
  • $\begingroup$ @AndréNicolas : Me, too. I wanted to give the OP some time to work on it on his/her own. $\endgroup$ – Christopher Carl Heckman May 14 '16 at 22:25
  • $\begingroup$ I will delete the comment to give OP time. The comment was made to correct the quite unreasonable official answer. $\endgroup$ – André Nicolas May 14 '16 at 22:27

I think the English of "some student asked every faculty member a question" is ambiguous. One could read it that there is a single student who asked every faculty member a question, or that every faculty member got asked a question by some student, but it could be various students. The book answer (with $\wedge$ instead of $\vee$ as you suggest) uses the second reading. Your answer is getting close to the first reading because starting with $\exists x$ forces it to be the same student doing the asking. You still need to work on the connectives. The difference between $\forall x \exists y Q(x,y)$ and $\exists y \forall x Q(x,y)$ is that in the first case we can find a different $y$ that works for each $x$, while in the second it has to be the same $y$ that works with every $x$.

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