Vacuous Truth and Universal Conditional Statements Sometime after I began studying conditional statements, I started having difficulty understanding vacuous truth. For instance, the fact that for any set $A$ we have $\emptyset\subset A$ is commonly justified by asking (almost) rhetorically, 'Well, can you find an element in the empty set that isn't in $A$?', which undoubtedly is what is required to show that the assertion is false, but that answer always caused me to naively ask, 'Well, can you find elements in the empty set that are in A?'  
It wasn't until I finally understood that the statement $\forall x\in A\,P(x)$ does not imply the existence of an $x$ for which $P(x)$ is true. It is only meant to imply that there is no $x$ in $A$ for which $P(x)$ fails (see page 14 here). It then becomes clear to see why the statement $\forall x\in\emptyset\,P(x)$ is always true, for this is the assertion that there is no $x$ in the empty set $\emptyset$ for which $P(x)$ fails, and since there are no elements in the empty set $\emptyset$, the assertion is true, vacuously.  
But, recently I learned that the two statements:  
(1) $\forall x\in A\,P(x)$
(2) $\forall x(x\in A\implies P(x))$  
have the same meaning. For $A=\emptyset$, (1) and (2) become: 
(1') $\forall x\in\emptyset\,P(x)$
(2') $\forall x(x\in\emptyset\implies P(x))$.  
As seen before, (1') is true. But it isn't clear to me why (2') is true. To explain this, someone might say, 'to claim that (2') is false is to say that $\exists x(x\in\emptyset\,\land\,\lnot P(x))$. Since there are no elements in the empty set, it is impossible to show that (2') is false.' Once again, I naively counter, 'since there are no elements in the empty set, how can we show that (2') is true?'  
I think that the source of my unwillingness to be satisfied with explanations that (2') is true is the lack of knowing the precise assertion that a conditional statement is making. As explained in the second paragraph, this certainly was the issue when trying to understand why (1') is true, but once I understood precisely the meaning of the statement $\forall x\in A\,P(x)$, it was easy to see why this was so. I think that if someone could explain precisely what (2') is asserting, specifically by addressing what is meant by a conditional statement, then it would be easy to see why (2') is true.  
UPDATE 1:
I should make something clear. When I wrote,  

But, recently I learned that the two statements:
  ...  

I meant that I became aware that statements (1) and (2) are equal but did not understand why. This led me to seek out information about universal statements and universal conditional statements. Now, I think I understand why (1) and (2) are equivalent. The universal conditional statement $\forall x\in U(P(x)\implies Q(x))$ means that every object $x$ in $U$ that satisfies $P(x)$ also satisfies $Q(x)$. If I'm not mistaken, this statement is equivalent to $\forall x\in D\,Q(x)$, where $D=\{x\in U:P(x)\}$.  
Also, after reading the answer from Graham Kemp, specifically where he writes,  

As you've seen, a universal statement is falsified if we can witness a counter example.   A vacuous truth is not falsified because no counter examples to any statement exist in the empty set.  

I considered (seriously) the negation of the statement $\forall x(x\in\emptyset\implies P(x))$, which is the statement $\exists x(x\in\emptyset\land\lnot P(x))$. The predicate, $x\in\emptyset$, is false for every possible value for x. Therefore, the existential statement can never be true. It follows that the original statement is true since its negation is false. This is (now) absolutely convincing that (2') is true.  
However, is this the explanation for the fact that for regular conditional statements, whenever the antecedent is false, the implication is true? Or, does the reasoning for this come from understanding the precise meaning of the conditional of two statements? The reason I ask this is because I read that when dealing with vacuous truth of universal statements, "we need predicate logic to be consistent with propositional logic" (see page 4 here) which suggests that it has already been established that a regular conditional statement is true when its antecedent is false.  
UPDATE 2:
This is in response to the last paragraph in UPDATE 1. In order to understand why the conditional of two statements is true when the antecedent is false, it is necessary to understand its precise meaning. The conditional statement, "if $p$, then $q$" is the assertion that "whenever $p$ is true, $q$ is also true." Thus, it is false only when $p$ is true and $q$ is false. In symbols:  
$\lnot(p\implies q)\equiv p\land\lnot q$  
Since a statement is true if and only if its negation is false, we have:  
$p\implies q\equiv\lnot(p\land\lnot q)$  
or,  
$p\implies q\equiv\lnot p\,\lor q$  
Thus, when $p$ is false (which means $\lnot p$ is true), $p\implies q$ is true, vacuously.  
UPDATE 3 (Fri Jan 15):
I'm having a bit of difficulty with the fact that any assertion about all of the elements of the empty set $\emptyset$ is vacuously true. For instance, I completely understand and agree with the fact that $\emptyset\subset A$ where $A$ is any nonempty set (i.e., $\forall x\in\emptyset :x\in A$ is a true statement). But isn't the statement $\forall x \in\emptyset : x\notin A$ also true?  
I guess my question is, can both of the statements $\forall x\in\emptyset : p(x)$ and $\forall x\in\emptyset : \lnot p(x)$ be true?  
RESPONSE TO ZeroXLR:
I guess what I'm seeking is how to confidently explain this to another person.  
I agree and understand that the negation of the statement (1) $\forall x\in\emptyset : p(x)$ is not (2) $\forall x\in\emptyset : \lnot p(x)$. The negation of (1) is the statement $\exists x\in\emptyset :\lnot p(x)$, which of course is never true. Hence, (1) is always true and that is how we show that the empty set $\emptyset$ is a subset of any set $A$.  
So, would I be correct in saying that since the statements (1) and (2) are both true, all the elements of the empty set satisfy both $p(x)$ and $\lnot p(x)$? And of course, if we take $p(x)$: $x\in A$, then this means that $\emptyset\subset A$.
 A: 
As seen before, (1') is true. But it isn't clear to me why (2') is true. 

These statements have the exact same meaning.   The first is simply a shortened expression.
$\forall x {\in}\varnothing\; P(x)$ "everything in the empty set satisfies the predicate"
$\forall x\; \big(x\in\varnothing \to P(x)\big)$ "if any thing is in the empty set, then that thing will satisfy the predicate".  
As you've seen, a universal statement is falsified if we can witness a counter example.   A vacuous truth is not falsified because no counter examples to any statement exist in the empty set.
This can also be explained as:   An implication is held to be true if either the antecedent is false or the consequent is true.   Because the antecedent ($x\in \varnothing$) is never true, therefore the implication always holds, irrespective of the consequent.   Thus no counterexamples to an implication of that form can ever exist.
A: 
This answer refers to your UPDATE 3 (Fri Jan 15):

When we say $x \in \emptyset \implies P(x)$ or $(\forall x \in \emptyset) P(x)$ is true an important thing to note is this: we are asserting that the entire statement is true. We are not asserting the truth of just $P(x)$ in isolation. This is why $(\forall x \in \emptyset) P(x)$ and $(\forall x \in \emptyset) \neg P(x)$ can both be true at the same time even though both $P(x)$ and $\neg P(x)$ can certainly not be true in classical logic.
To further help you clear your confusion, I consider the case when $P(x) \equiv x \in A$ as in your question. You might still be puzzled by the fact that if $(\forall x \in \emptyset) x \not\in A$ is also true as I just said, how can we still claim that $\emptyset \subseteq A$? This is where you have to remember to consider the statement $(\forall x \in \emptyset) x \not\in A$ as a whole: the negation of $(\forall x \in \emptyset) x \in A$ (i.e. $\emptyset \subseteq A$) is not $(\forall x \in \emptyset) x \not\in A$. It is $(\exists x \in \emptyset) x \not\in A$ which is a subtly different statement: you need to demonstrate the existence of an $x$ that is simultaneously in $\emptyset$ but not in $A$. But of course, you cannot do that!

EDIT: Response to your response:

Yes, all elements of the empty set (of which there are none!) do indeed satisfy $p(x)$ and $\neg p(x)$ for any proposition $p(x)$, where $x$ is a dummy variable. This would have been contradictory if only we could concretely identify one such $x$. But since we can't, $x$ will remain a dummy variable that refers to elements that do not actually exist.
On a slightly unrelated but illuminating note, I would like to say one last thing: Confusions such as these arise, in general, when we mix up the usual meanings of phrases like "for all" and "there exists" in natural language with the same phrases in math. The meanings are undoubtedly related but there are subtle differences like the case here. In normal English (unless you are talking to a mathematician or a lawyer), when we say "for all something...", we usually implicitly assume the existence of at least one of that something. Failure to assume that might result in people thinking that you are trying to trick them! In math, however, the case is different.
A: Answering your last question.  Yes, they are both true.  In fact, $\forall{}x\in\emptyset:1=2$ is true.
Stepping back a bit though, I prefer constructive mathematics and in particular find the BHK interpretation (or the general thrust of it) to be a good way of being comfortable with many logical results.  (And it has the benefit of directly connecting to computation which strongly grounds them.)
In the BHK interpretation, a proof of a statement $\forall x\in A: P(x)$ is a function that takes an element $x$ of $A$ and produces a proof of $P(x)$.  Similarly, a proof of a statement $P \implies Q$ is a function that takes a proof of $P$ and produces a proof of $Q$.  For context, a proof of $P \land Q$ is a pair of a proof of $P$ and a proof of $Q$, so the theorem $P \land Q \implies P$ is witnessed by the function $f(p,q) = p$.  Both of the statements you asked about are witnessed by functions from the empty set, i.e. $f : \emptyset \to P$ and there is always exactly one function of that type, namely the empty function.  In the latter question, via this interpretation, this is explicitly what you are asking for.  In the former question, it's a little more roundabout since you have a function $f : x \in \emptyset \to P$ but since $x \in \emptyset$ is false, i.e. $x \in \emptyset \implies \bot$ ($\bot$ being falsity), we can reduce this to $g : \bot \to P$ and, by definition, the set of proofs of $\bot$ is empty.  Indeed, to reconstruct the overall proof we just compose, $f = g \circ h$ where $h : x \in \emptyset \to \bot$ is the proof that $x \in \emptyset$ is always false.
To reiterate, this interpretation validates constructive, but not classical, logic and so some classical results are not provable (in particular, any involving excluded middle).  Nevertheless, for the things provable in this interpretation, the proofs are much more concrete than pure symbol manipulation; you can see how the pieces fit together.  (The proofs are also code; you can run them.)
