Assume the statement is false. Then the hypothesis ($\forall{x}.A(x)\rightarrow B(x))$ is true and the conclusion ($\forall{x}.A(x)\rightarrow \forall{x}.B(x))$ is false. Since the conclusion is false, its hypothesis ($\forall{x}.A(x)$) is true and its conclusion ($\forall{x}.B(x)$) is false. So there exists $c$ such that $\neg B(c)$. Specializing the other two universal quantifiers, we have $A(c)\rightarrow B(c)$ and $A(c)$, from which we deduce $B(c)$, bringing us to the desired contradiction.
Your (1) would be something to prove in whatever formalization of propositional or first-order logic you're using; most of the formal proof systems I'm familiar with don't have the contrapositive principle directly stated as an axiom, but it follows from the other axioms.
Then, the argument in (2) and (3) is fairly directly in the style of proof that comes under a "natural deduction" type of formal proof system. In these systems, if you can start with a hypothesis $p$ which you assume to be true and reach a conclusion $q$, then a rule of the formal proof system, often known as "$\rightarrow$-introduction", will allow you to "abstract" or "encapsulate" that into a proof of the implication $p \rightarrow q$. Thus, the form of your argument would look like:
\begin{align*}
(A \rightarrow B) & \vdash (\lnot B \rightarrow \lnot A) \\
\hline
& \vdash [(A \rightarrow B) \rightarrow (\lnot B \rightarrow \lnot A)].
\end{align*}
(In Hilbert-style proof systems, we similarly have the Deduction Theorem which fills a similar role.)
Now, as to how you would actually prove (1), that depends more on the details of the formal proof system. In the system I personally prefer to use, $\lnot A$ is defined as being the same as $A \rightarrow \bot$ where $\bot$ is an atomic formula representing a "false" proposition. (Think of the humorous way we sometimes use to negate a statement: instead of saying just "it's not raining," somebody might say "if it's raining, then I'm a frog.") Then the overall proof might look something like:
- $A \rightarrow B$ (assumption)
- $\quad \lnot B$ (assumption)
- $\quad \quad A$ (assumption)
- $\quad \quad \quad B$ (modus ponens from 1 and 3)
- $\quad \quad \quad \bot$ (modus ponens from 2 and 4)
- $\quad \quad A \rightarrow \bot$, or in other words $\lnot A$ ($\rightarrow$-intro from 3 through 5)
- $\quad \lnot B \rightarrow \lnot A$ ($\rightarrow$-intro from 2 through 6)
- $(A \rightarrow B) \rightarrow (\lnot B \rightarrow \lnot A)$ ($\rightarrow$-intro from 1 through 7)
(Here, the indentation is used to indicate that "assumption" lines above the current line, with a smaller amount of indentation, are part of the current proof context of what we are assuming to be true.)
Another way of representing the same proof would be as a proof tree:
$$ \frac{
\frac{
\displaystyle \frac{
\frac{
\displaystyle A \rightarrow B, \lnot B, A \vdash B \rightarrow \bot \qquad
\frac{
\displaystyle A \rightarrow B, \lnot B, A \vdash A \rightarrow B \qquad A \rightarrow B, \lnot B, A \vdash A}{\displaystyle A \rightarrow B, \lnot B, A \vdash B}
}{\displaystyle A \rightarrow B, \lnot B, A \vdash \bot}
}{A \rightarrow B, \lnot B \vdash \lnot A}
}{\displaystyle A \rightarrow B \vdash \lnot B \rightarrow \lnot A}
}{\vdash (A \rightarrow B) \rightarrow (\lnot B \rightarrow \lnot A)} $$
This representation has the advantage that it makes it clear at each element of the proof what propositions are (provisionally) being assumed to be true, and what is being concluded. (It obviously has the disadvantage of taking up a great deal of space, though.)
Best Answer
To prove that two formulas $\phi$ and $\psi$ are logically equivalent, you need to show that 1) $\phi$ logically entails $\psi$ and that 2) $\psi$ logically entails $\phi$.
To show that some formula $\phi$ entails some formula $\psi$, you need to show that all structures where $\phi$ is true $\psi$ is true as well, that is, that there is no situation where $\phi$ is true but $\psi$ false.
To show that two formulas are not equivalent, you need to show that at least one of the entailment directions does not hold.
To show that one formula does not entail the other, you need to show that there is at least one structure in which the first formula is true but the second is not.
The first thing for you to find out is whether or not the two formulas actually are logically equivalent. To make things shorter, I'll just spoiler you that they are not.
So in order to prove that the two formulas are not equivalent, we need to show that one of the entailment directions doesn't hold, i.e. either
i.e. there is at least one structure where $∃xP(x)⟹∃xQ(x)$ is true but $∃x[P(x)⟹Q(x)]$ false
or
i.e. there is at least one structure where $∃x[P(x)⟹Q(x)]$ is true but $∃xP(x)⟹∃xQ(x)$ false.
Let's pick the second option. We need to find a structure in which $∃x[P(x)⟹Q(x)]$ is true but $∃xP(x)⟹∃xQ(x)$ false. It suffices to find one such counterexample and we showed that the entailment, and hence the logical equivalence, does not hold in general. Let's assume a simple structure with two elements $\{a, b\}$, and an interpretation such that $P$ is true of $a$ but not $b$, and $Q$ is false for both $a$ and $b$:
You were already pretty close with your approach. The only thing you need to change is to make $Q$ false of the second element, so $\exists x Q(x)$ and thereby $∃xP(x)⟹∃xQ(x)$ becomes false, while $\exists x P(x)$ is still true.
Now since for the element $b$ we have that $P(b)$ and $Q(b)$ are both false, then by the truth table of $⟹$, the implication $P(x)⟹Q(x)$ is true for that element $b$, and since we found an element $x$ that fulfills the inner formula, we can establish $∃x[P(x)⟹Q(x)]$ is also true.
Now we know that $∃x[P(x)⟹Q(x)]$ is true and we need to show that $∃xP(x)⟹∃xQ(x)$ is false. (Reminder: This is because we want to show that from $∃xP(x)⟹∃xQ(x)$ being true it doesn't necessarily follow that $∃x[P(x)⟹Q(x)]$ is true, so we need to find one assignment which is a counterexample to the validity of the entailment.)
Remember that we assumed that there is at least one element of which $P(x)$ holds -- here we called that element $a$ -- but that $Q$ is false for every element. Now with $a$ we found an element $x$ such that $P(x)$ is true, so $\exists P(x)$ is true, but there is no $x$ with $Q(x)$, so $\exists x Q(x)$ is false. Now we have that the left-hand-side of the implication is true but the right-hand side is false. By the truth table of implication, this makes the statement $∃xP(x)⟹∃xQ(x)$ false.
So we found an interpretation under which $∃x[P(x)⟹Q(x)]$ is true but $∃xP(x)⟹∃xQ(x)$ is false. So $∃x[P(x)⟹Q(x)]$ does not logically entail $∃xP(x)⟹∃xQ(x)$. Hence, the two formulas can not be equivalent.
TL;DR: To prove that two formulas are not equivalent, you need to show that either the one doesn't entail the other or the other doesn't entail the one. To show that an entailment does not hold, you need to find a structure where the first formula is true but the second one is not. To show that an implication is false, you need to show that the left-hand side is true but the right-hand side is false.
If the two formulas were indeed equivalent, you would need to show that they logically entail each other, i.e. you'd need to argue that no matter the structure, if one formula is true, then the other must be true as well and vice versa.