Source: p. 69. How to Prove It by Daniel Velleman. I already read 1,2, 3, 4, 5, 6, 7, 8, 9, 10, 11 & 12.
$\exists \, x \, \in \, \emptyset : P(x) \tag{1}$ will be false no matter what the
statement $P(x)$ is. There can be nothing in $\emptyset$ that, when plugged in for $x$, makes $P(x)$
come out true, because there is nothing in $\emptyset$ at all! It may not be so clear whether $\forall \, x \, \in \,\emptyset : P(x) $ should be considered true or false …
I then paused reading to conjecture the truth value of: $2. \; \forall \, x \, \in \,\emptyset : P(x). $
$\boxed{\text{Conjecture :}}$ Because $\emptyset$ contains nothing, ergo $x \, \in \,\emptyset$ is false $
\implies \forall \, x \, \in \,\emptyset $ is false.
Since the Truth Value of $P(x)$ is unknown, ergo 2 is false. $\blacksquare$
Then I continued reading and was stupefied to learn that 2 is true:
After expanding the abbreviation of the quantifiers, $\forall \, x \, \in \,\emptyset : P(x) \quad \equiv \quad \forall \, x \, \left[\, x \, \in \,\emptyset \Longrightarrow P(x)\right]. \tag{*}$
Now according to the truth table for the conditional connective, the
only way this can be false is if there is some value of $x$ such that $x \, \in \,\emptyset $ is true but $P(x)$ is false. But there is no such value of $x$, simply because there isn’t a value of $x $ for which $x \, \in \,\emptyset $ is
true.
Thus, (*) is (vacuously) true.
Though I understand this proof and the Principle of Explosion, I still do not understand why my intuition failed. How can I correct my intuition?
Supplement to mercio's Answer
I understand $\forall x \in \emptyset,P(x). \; \stackrel{dfn}{\equiv} \; \forall x, \color{#B22222}{x\in \emptyset}\implies P(x). \quad \equiv \; \forall x,\color{#B22222}{false}\implies P(x)$.
Consider $3. \forall\;\bbox[5px,border:2px solid #32CD32]{\, \color{#B22222}{\underbrace{{x\in\emptyset}}_{false}} \;,P(x)} \;$. 3. Is 3 vacuously true because the green box is vacuously true?
I consider the green box above a statement, because though the comma is not a Logical Connective,
$ \forall \; \color{#B22222}{\underbrace{{x\in\emptyset}}_{false}} \; \;P(x) \quad \equiv \quad \forall \; \color{#B22222}{\underbrace{{x\in\emptyset}}_{false}} \; \color{#0073CF}{\huge{,}} \;P(x) \quad \equiv \quad \forall \; \color{#B22222}{\underbrace{{x\in\emptyset}}_{false}} \; \color{#0073CF}{{\huge{\text{,}}} \text{we have that}} \; P(x). \tag{**}$
Best Answer
You are right when you say that $\forall x$, the statement $x \in \emptyset$ is false. This means that $\forall x, \neg (x \in \emptyset)$, which is equivalent to $\forall x, x \notin \emptyset$. Perfectly true.
Then you say "the statement $\forall x \in \emptyset$ is false". $\forall x \in \emptyset$ is NOT a statement, it's an incomplete sentence. Either you write "$\forall x, P(x)$", either you write "$\forall x \in X, P(x)$", which is a shorthand for "$\forall x, (x \in X \implies P(x))$". "$\forall x \in \emptyset$" is not a statement. It can't be true or false.
$\forall x \in \emptyset, P(x)$ is a shorthand for $\forall x, (x \in \emptyset \implies P(x))$, which is equivalent (since $x \in \emptyset$ is always false) to $\forall x, ~\textrm{false} \implies P(x)$. After looking at the truth table for $\implies$, this is equivalent to $\forall x, ~\textrm{true}$ (whatever $P(x)$ may be), which is $\textrm{true}\;.$
If you want to disprove $\forall x \in \emptyset, P(x)$ you have to show me an $x \in \emptyset$ such that $P(x)$ is false. Well you will never find an $x \in \emptyset$.