Consider the first-order sentence
(1) $\forall x\exists y(\forall z Dxz\to\exists z\neg Dyz)$
and interpret Dab as the two-place relation "a has doubts about b." On a recent exam, I translated the sentence back into English as
(2) For everyone who has doubts about everything, there's someone who doesn't have doubts about something.
My instructor counted off & instead gave the answer as:
(3) Even if there is someone who has doubts about everything, there is still someone who does not have doubts about something.
But I can't see how these two translations come apart -- i.e. I can't think of a model where one is true but the other is false. What am I missing?
Let me elaborate just a bit. (3) is certainly the best translation of (1) after it has been transformed into what Quine called "pure form": by the laws of quantifier passage, (1) is equivalent to
(1') $\exists x\forall z Dxz\rightarrow \exists y\exists z\lnot Dyz$
But with (2) I was trying to expose the general dependency, in the manner of Skolem functions, of the existentially quantified y on the universally quantified x. I grant that there's a reading of (2) according to which I'm asserting that for every person who has doubts about everything there's a unique person who doesn't have doubts about something (a much stronger claim than (1)), but it seems to me that (2), on the face of it, doesn't require this reading -- it doesn't say outright that y=f(x) is injective and indeed is consistent with it being constant.
So my question is really aimed at logic instructors. Would you have counted off if I offered this translation, and if so, why? The best explanation I can see is that (2) is ambiguous with its most natural reading being the stronger reading I explained above. Or if I really am missing something here, please point it out.