Quantified logic: is my answer equivalent to the other solution? I'm reviewing for an exam, and I'm having trouble with a practice problem.
Translate the following into quantified logic: There is a student who sells cookies and another student with straight A's whenever all students are home for winter break.
Let C(x) = "x sells cookies", A(x) = "x has straight A's", and W(x) = "x is from for winter break".
The solution given to me is: $\forall x(W(x) \rightarrow \exists y \exists z(C(y) \land A(z)))$
I think I see how that is correct, but I came up with a different answer and was wondering if it is equivalent to that solution.
My solution: $\forall x(W(x) \rightarrow (\exists y C(y) \ \land \ \exists z A(z)))$
Would my solution be correct too?
 A: Let $S$ be the given sentence.  $S$ has the form "$P$ whenever $Q$," where $P$ is 
$$\exists y \exists z (C(y) \wedge A(z) \wedge y \neq z)$$
and $Q$ is $\forall x\, W(x)$.  Combining $P$ and $Q$ into $Q \rightarrow P$, we get
$$(\forall x \, W(x)) \rightarrow \exists y \, \exists z \, (C(y) \wedge A(z) \wedge y \neq z) \enspace.$$
Putting this sentence in prenex form, we get
$$ \exists x\, \exists y\, \exists z \,(W(x) \rightarrow (C(y) \wedge A(z) \wedge y \neq z)) \enspace. $$
Let's translate it back into English: there are three students, $x$, $y$, $z$, the last two distinct, such that either $x$ is not home for winter break, or $y$ sells cookies and $z$ has straight A's.  We can make this sentence true by exhibiting a student who's not home for the break, because then we are under no obligation to show two more students to satisfy the consequent of the implication; or we can make the sentence true by exhibiting two distinct students, one selling cookies and the other having straight A's, because then the fact that all students are home for the break is of no concern. 
Compare the sentence above to
$$ \forall x\, \exists y\, \exists z \,(W(x) \rightarrow (C(y) \wedge A(z) \wedge y \neq z)) \enspace. $$
This sentence is false if at least one student is home for the break, but no student sells cookies (or no student has straight A's, or the only student with straight A's is also the only one to sell cookies).  As far as I can tell, that's not what $S$ claims.
