The answer is indeed A, but your reasoning is a bit off: specifically, quantification over the emptyset is irrelevant here (although I think your intuition is coming from the right place). Also, you haven't used the assumption that the language has no function symbols, which is crucial.
- Can you construct a counterexample if we allow function symbols? HINT: What does the sentence "$\forall x(x\not=f(x)\wedge x\not=f(f(x)))$" imply about the size of the structure? Do you see how to generalize this?
Rather, you can argue as follows:
Let $M\models\varphi$. (It doesn't matter how big a model we pick, here.)
Since $M\models\varphi$ we get witnesses $a,b,c\in M$ - possibly not distinct - such that $M\models\forall v,w,x,y[\psi(a,b,c,v,w,x,y)]$.
Because the language of $M$ has no function symbols, $N=\{a,b,c\}$ forms a substructure of $M$. Now, do you see why we have $N\models\forall v,w,x,y[\psi(a,b,c,v,w,x,y)]$ (and hence $N\models\varphi$) as well?
- HINT: remember that $\psi$ has no quantifiers, and by cutting down to a smaller structure we only restrict the possible values for the universal quantifiers - there's a general fact here about how sentences of the form [universal quantifiers][quantifier-free part] behave with respect to taking substructures ...
First of all, please know that these are not equivalences in the strict sense of logical equivalence ... with $n$ terms in $P(x_1) \land ... \land P(x_n)$ the best we can say is that that statement will have the same truth-value as $\forall x \ P(x)$ for any domain with $n$ elements, and where $x_1$, $x_2$ ... are treated as constants that respectively denote each of those $n$ different objects. Indeed, to make that clear, I would use $c_i$'s rather than $x_i$'s
Still, as long as you are careful and understand this restriction, you can indeed usefully work with this 'equivalence' ... which I'll denote by $\approx$
Now to your question. If you have multiple quantifiers you can just work them out step by step:
$$\exists x \forall y \ P(x,y)\approx$$
$$\forall y \ P(c_1,y) \lor \forall y \ P(c_2,y) \lor .... \lor \forall y \ P(c_n,y) \approx$$
$$(P(c_1,c_1) \land P(c_1,c_2) \land ...P(c_1,c_n)) \lor (P(c_2,c_1) \land P(c_2,c_2) \land ...P(c_2,c_n)) \land .... \land (P(c_n,c_1) \land P(c_n,c_2) \land ...P(c_n,c_n))$$
You can also work this out inside out:
$$\exists x \forall y \ P(x,y)\approx$$
$$\exists x (P(x,c_1) \land P(x,c_2)\land ... \land P(x,c_n)\approx$$
$$(P(c_1,c_1) \land P(c_1,c_2) \land ...P(c_1,c_n)) \lor (P(c_2,c_1) \land P(c_2,c_2) \land ...P(c_2,c_n)) \land .... \land (P(c_n,c_1) \land P(c_n,c_2) \land ...P(c_n,c_n))$$
Best Answer
You shouldn't really think of "three quantifiers" as a separate case from "two quantifiers". Think of it one quantifier at a time.
$\exists x \varphi(x)$ means that there is some $x$ that makes $\varphi(x)$ true.
$\forall y \exists x \varphi(x,y)$ means that no matter which $y$ you pick, $\exists x\varphi(x,y)$ is true; so no matter which $y$ you pick, there is some $x$ that makes $\varphi(x,y)$ true.
$\exists z\forall y\exists x\varphi(x,y,z)$ means that there is some particular $z$ so that the statement $\forall y\exists x\varphi(x,y,z)$ is true. So there is some particular $z$ so that no matter which $y$ you pick, there is some $x$ that makes $\varphi(x,y,z)$ true.
You can tell what depends on other things based on the order: in this last example, $y$ depends on nothing, but the choice of $x$ may depend on both $y$ and $z$. As a general rule, variables bound by $\forall$ quantifiers don't depend on anything, because they aren't being "chosen" (you're talking about all possibilities for them). Variables bound by $\exists$ quantifiers depend on all variables that preceded them.
I like to think of heavily-quantified statements as games, played between me and an opponent. I'm trying to make my statement true, while my opponent is trying to make it false. Whenever we see an $\exists$ quantifier, I get to pick the value of the variable; whenever we see an $\forall$, my opponent gets to pick. The statement is true if I have a winning strategy - that is, if there's a way for me to win no matter how my opponent plays.
For example, say we're looking at the sentence $\forall x\exists y\forall z(z \leq x \vee y \leq z)$, and all quantifiers are over the set of natural numbers $\mathbb{N}$. This statement is true, and here's why: first, my opponent gets to pick $x$. Whatever $x$ he picked, I'll choose $y = x + 1$. Now, no matter which $z$ he picks, $z$ will be either at most $x$ or at least $x + 1$ (because there aren't any numbers between $x$ and $x + 1$). Because I can always win this way, the original statement is true. And because we saw why this is a winning strategy, you can also see what the original statement was "really" saying: "for every $x$, there is a $y$ so that nothing is bigger than $x$ but smaller than $y$"; or, put more succinctly, "every number has a next number".