Logical equivalence of quantified propositions involving if-then

Question

Is

$$\exists x V(x) \rightarrow \forall cS(c)$$

logically equivalent to

$$\exists x (V(x) \rightarrow \forall cS(c))?$$

Is it enough to say that they are NOT logically equivalent simply because the "outside structure" of the first proportion is "If ... , then ..." while that of the second structure is "There exists ..."?

Answer 1

Concerning your first question: No, they are not equivalent. Consider the structure $$\mathfrak{A} = \langle \mathcal{A}, \mathcal{I} \rangle$$ with $$\mathcal{A} = \{a,b\}$$ $$\mathcal{I}(V) = \{a\};\\ \mathcal{I}(S) = \emptyset$$

$\mathfrak{A} \models \exists x (V(x) \to \forall c S(c))$, because with $b$, of which $V$ doesn't hold, the antecedent of the implication is false, so by definition of $\to$, the implication becomes true, and since we found one $x$ such that $V(x) \to \forall c S(c)$ is true, the existential quantification is true.
But $\mathfrak{A} \not \models \exists x V(x) \to \forall c S(c)$: With $a$, there is an element from the domain of which $V$ holds, which makes the existential quantification and hence the antecedent of the implication true, so by the definition of $\to$, the consequent must be true as well; but with $a$ (and $b$) there are individuals of which $S$ does not hold, so the univeral quantification and hence the consequent of the implication is false, so by the definition of $\to$, the statement is false.
Since there is a model that satisfies one but not the other formula, the formulas can not be equivalent.

$\exists x (V(x) \to \forall c S(c))$ says "There is something such that if this something is V, then everything is S".
$\exists x V(x) \to \forall c S(c)$ says "If there is a V, then everything is S".
These two statemens are not equivalent.

If you find this confusing, you might be interested to read about the drinker paradox.

In general, we have the following logical equivalences, provided that $x$ does not occur free in $\psi$ (otherwise, by moving the quantifier out, we would accidentally bind the variable $x$ in $\psi$, which changes its interpretation):

$(\psi \to \exists x \phi) \equiv \exists x(\psi \to \phi)$
$(\psi \to \forall x \phi) \equiv \forall x(\psi \to \phi)$
$(\exists x \phi \to \psi) \equiv \forall x(\phi \to \psi)$
$(\forall x \phi \to \psi) \equiv \exists x(\phi \to \psi)$

That is, when moving quantifiers out of the inside an implication to quantify over the implication as a whole, if they were initially in consequent position ($\psi \to Q\phi$), they stay the same, and if they were initially in antecedent position ($Q\phi \to \psi$), they change to their dual ($\exists$ becomes $\forall$, and $\forall$ becomes $\exists$).
So an equivalent statement to $\exists x V(x) \to \forall c S(c)$ would be $\forall x (V(x) \to \forall c S(c))$.
For the respective equivalences of moving quantifiers in and out of other connectives, you can read up the rules of how to transform a formula into prenex normal form.

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Concerning your second question: No, from the fact that two formulas are syntactically not identical we can not conclude that they are not logically equivalent.
For example, $(\exists x V(x) \to \forall c S(c)) \equiv (\exists x V(x) \to \forall c S(c)) \lor (\exists x V(x) \to \forall c S(c))$.
Or $(\exists x V(x) \to \forall c S(c)) \equiv (\exists x V(x) \to \forall c S(c)) \land (\forall x (x=x))$.
Or $(\exists x V(x) \to \forall c S(c)) \equiv \forall x (V(x) \to \forall c S(c))$.
Or, as you figured out, $\forall x P(x) \land \forall x Q(x) \equiv \forall x (P(x) \land Q(x))$.
For any formula, there are infinitely many formulas to which the formula is logically equivalent, some of them are trivial, some of them more interesting (e.g. the equivalence $\exists x V(x) \to \forall c S(c) \equiv \forall x (V(x) \to \forall c S(c))$.
In general, you have a many-to-one mapping from syntax to semantics: One syntactically unique formula can not have more than one meaning; but there can in principle be more than one syntacticlly different forms that have the same semantics.

Answer 2

Putting in brackets to clarify order of operations. $$\exists x~\Big(V(x)\to\big(\forall c~S(c)\big)\Big)\\\text{versus}\\\big(\exists x~V(x)\big)\to\big(\forall c~S(c)\big)$$

Consider the universe of integers and the predicates $V(x): x\text{ is odd}$, and $S(c):c\text{ is even}$.

Therefore $\forall c~S(c)$ is false; since not every integer is even.

$~$

Now $V(0)\to\big(\forall c~S(c)\big)$ will be true, because $0$ is not odd thus making the conditional true.   Thus it is the case that "Thereis some integer such that every integer is even if that integer is odd".   It is counter-intuitive, but verified.

$~$

However, $(\exists x~V(x))\to\big(\forall c~S(c)\big)$ is false because there does exist an odd integer.   So it is clearly not the case that "Every integer is even if there is some integer that is odd."

So in that interpretation we have: $\exists x~\Big(V(x)\to\big(\forall c~S(c)\big)\Big)$ is true while $\big(\exists x~V(x)\big)\to\big(\forall c~S(c)\big)$ is false.

Thus demonstrating that the statements are not logically equivalent.