translating informal language to predicate logic; existential vs universal quantifier I am analysing some problems in the Hurley Logic Text book (12th edition) and I confused about a couple of answers he gives. 
Here are two related statements that he translates differently (taken from Hurly p. 505). The first with a universal quantifier and the second with an existential quantifier
1) "If there are any safe drivers, then if none of the trucks break down they will be hired.
2) "If there are any safe drivers, then some safe drivers will be hired."
Hurley's translation: 
(Sx: x is safe, Dx: x is a driver, Tx: x is a truck; Bx: x breaks down; Hx: x will be hired)
1) $(\forall x)[(Sx \wedge Dx) \rightarrow [(\forall y)(Ty \rightarrow \neg By) \rightarrow Hx]]$
2) $(\exists x)(Sx \wedge Dx) \rightarrow (\exists x)[(Sx \wedge Dx) \wedge Hx]$
My sense is that, if this is not an error, Hurley is trying to stress something in the first sentence that should trigger a universal quantifier, even though the identical phrase "if there are safe drives"  is used in the second sentence where the existential quantifier is used (presumably correctly).
However I don't see it. Intuitively, I would have written the first statement as:
1a) $(\exists x)[[(Sx \wedge Dx) \wedge (\forall y)(Ty \rightarrow \neg By)] \rightarrow Hx$]
Which to me says, if there exists some safe drivers (assuming none of the trucks break down) then some safe drivers will be hired. 
Hurley's translation strikes me as saying: "everything that is a safe drive (assuming none of the trucks break down) will be hired". 
I wonder if anyone can explain where I'm going wrong. Thanks
 A: Please note that any conditional is true as soon as its antecedent (the 'if' part) is false.
So, we can easily make your:
$(\exists x)[[(Sx \wedge Dx) \wedge (\forall y)(Ty \rightarrow \neg By)] \rightarrow Hx]$
true by pointing to any one object that is not a driver, or that is not safe.  Put differently: this statement would be true in any domain where there is at least one non-driver, and anything the statement is trying to say about the actual safe drivers has become completely irrelevant.  So, this is certainly not what you want.
It is indeed the reason why you almost never see an existential claim being an existentially quantified statement of a conditional: they can almost always be made vacuously true. 
OK, so why use a universal in the case of 1?  Because the use of 'a' is really reflecintg 'any.  That is, it is saying: "for anything at all, if it is a safe driver, then ..."
Consider this sentence:  "If a cat whips its tail, the it is mad".  Here, we clearly make a universal statement about any cat, and so this should be symbolized as something like:
$\forall x ((C(x) \land W(x)) \to M(x))$
Again, the existential version of this:
$\exists x ((C(x) \land W(x)) \to M(x))$
can be made vacuously true by something that is not a cat, and hence this latter statement does not end up forcing, and thus saying anything, about an actual cat that whips its tail.
OK, but then why in 2 do we use an existential?  It is because 2 is really of a different form. Statement 2 is in fact not a quantified statement at all!  It is a conditional ... whose antecedent and consequent both happen to be quantified statements.  Most importantly, the 'then' part is not making any references to any of the objects referred to in the first statement, so you don't have one quantifier quantifying over a conditional where the 'if'; and 'then' part have the same subject,  but rather a conditional relating two quantified statements, each with their own subject.
So, 1 and 2 are quite different statements! But when you express them in English, they sounds very similar. Indeed, statements of type 1 are often referred to as 'donkey sentences', after the famous:
"If a farmer owns a donkey, then he will beat it"
Do you see now why this sentence is a (double) universal rather than a (double) existential?
