How can I correct my wrong Intuition that $\forall \, x \, \in \,\emptyset : P(x) \quad $ is false?

_{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{**}$

Solution 1:

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 $x \in \emptyset$.

Solution 2:

IMO, it's our grasp of natural language that leads us astray here. Natural language is not about explicitly expressing precise ideas; there is a lot of ambiguity and implicit inference involved.

In particular, one doesn't speak about "all of something" unless there is (possibly hypothetically) actually something to speak about.

As you are trained to make this inference, when you hear "$\forall x \in \varnothing:P$", you mentally add the implicit hypothesis "$\exists x \in \varnothing$", which is where your intuition goes awry.