To be honest, Rosen's book is not reliable when it comes to logical niceties. For example, in the passage you are referring to, he writes
The statement βx is greater than 3β has two parts. The first part, the variable x, is the subject
of the statement. The second partβthe predicate, βis greater than 3ββrefers to a property that
the subject of the statement can have.
For a start, βx is greater than 3β isn't a statement in a normal sense of the term (it has a dangling free variable), and -- it we take subject terms in the ordinary sense to be putative referring expressions -- then βxβ isn't a subject term either.
As for predicates, how best to think of them (as strings of words, as strings of words with marked gaps, or other alternatives) has been a vexed question ever since Frege. There's a lot initially to be said for the Fregean line that we should treat predicates in the second way -- so for example '$\xi$ killed $\zeta$' and '$\xi$ killed $\xi$' count as distinct predicates (the first murderous, the second suidical!), with '$\xi$' and '$\zeta$' as place markers. But the arguments about the best treatment soon get murky: see this fine article http://www.phil.cam.ac.uk/teaching_staff/oliver/What_is_a_Predicate.pdf
Some comments:
(i) In propositional logic, a propositional variable stands for a sentence, i.e. something that can be true or false.
This is the reason why, starting from a first-order formula, we have to "remove" free variables; a well-formed f-o formula like "$x$ is green" (i.e. $green(x)$) is not a sentence, because we cannot say, without assigning a reference to $x$ if it is true or false.
Thus, we have to do one of the two thing mentioned by you: either replace $x$ with a constant ( a "name") or bind the variable with a quantifier.
(ii) $\forall x green(x)$ is a prefectly understandable sentence: "all objects are green".
You are right in saying that in a finite domain, $\forall x green(x)$ is equivalent to : $green(c_1) \land \ldots \land green(c_n)$.
But with first-order logic we are interested to "handle" also infinite domain; and in mathematical logic, we are interested primarily to handle with infinite domain.
And we do not (usually) admit as well-formed formulae infinite long expressions (but see Infinitary Logic).
If we restrict ourselves to finite domain, an universally quantified formula is a (finite) conjunction, while an existetially quantified formula is a (finite) disjunction.
Thus, if we restrict ourselves to finite domain, we can simply dispense with quantifiers ...
(iii) Having said that, what is the purpose of "downsizing" a first-order formula to a propositional one?
This technique can be useful in order to show that a f-o formula is valid: if the "propositional translation" of a f-o formula is a tautology, we are sure that the original f-o formula is valid.
Consider for example the (obviously) valid formula:
$\forall x (x=0) \lor \lnot \forall x (x=0)$;
we can translate it as $p \lor \lnot p$, that is a tautology.
Thus the original f-o formula is valid.
But we have valid formulae that are not "instances" of tautologies; consider :
$\forall x (x \ge 0) \rightarrow (0 \ge 0)$.
Its "translation" is $p \rightarrow q$, that is not a tautology.
This fact must be obvious; first-order language as a greater "expressive power" than propositional language; with f-o language we can perform a "deeper" decomposition of the logical structure of sentences (compared to the "decomposition" that we can perform only in term of the conncetives).
Thus, we are able to "discover" more logical truth than with "propositional decomposition".
Best Answer
I hear a few potential misunderstandings based on the title of this question compared to the full question.
First is that when we say that the domain of $P(x)$ is the odd numbers between $-6$ and $6$, we are trying to determine for which of those values $P(x)$ is true, and which ones are false. Phrased in a more intuitive human way, it's asking: "In the eyes of the property $P$ which values of $x$ feel right and true, and which ones are wrong and false"
This is often phrased in math circles as "for what values of $x$ does the property $P$ hold", which is short hand for "for what values of $x$ does $P$ hold true"
So in this case when we write $P(-5)$ we are saying (in the more human phrasing) that $P$ has determined that $-5$ is true. In math terms we would say that "$P$ holds for $-5$." And when we write $\neg P(-5)$ we are saying (again first in the more human way) that $P$ has determined that $-5$ is false. In math terms we would say "$P$ does not hold for $-5$.
We cannot have a situation where $P$ both holds and does not hold for a single value of $x$. $P$ can not determine that $-5$ is both true and false at the same time. $P(x) \land\neg P(x)$ is a contradiction.
In the answer you provided if we are to translate it into English words you are saying: The property $P$ holds for the entire domain (this is the second half of your answer). But there is also a number in the domain that doesn't hold (this is the first half). This implies that for at least one of the numbers: $-5, -3, -1, 1, 3, 5$ it is the case that $P$ thinks it's both true and false, which we just said can't happen.
Let's translate the statement in the question into English. First we have $\exists x(\neg P(x))$. This is saying that there exists some $x$ in your domain where the property $P$ does not hold.
The next part which is connected by a conjunction is $\forall x((x < 0) \to P(x))$ which is saying that "for all elements in your domain, it is the case that if $x$ is less than zero, then the property $P$ holds." In other words: "$P$ holds for all of the elements in your domain less than $0$"
As you figured out, you need to give an answer that only has $P(x)$, $\neg P(x)$, $\lor$, $\land$, and parentheses, for each specific number in the domain. However the problem at hand is writing something that is consistent with the statements "There exists an $x$ such that $P$ does not hold" and "$P$ holds for all $x$ less than $0$".
I hope that helps!