Is $ \forall x \in \emptyset. (P(x) \land \lnot P(x)) $ true? I have searched the forums but haven't found any answers to this question, so I thought I'd ask it myself.
Let $P(x)$ be an arbitrary predicate. Does the following statement evaluate to true or false?
$ \forall x \in \emptyset. (P(x) \land \lnot P(x)) $
It should evaluate to true because there is no element, hence it is true for all elements.
But it should evaluate to false because $ P(x) \land \lnot P(x) $ is always false.
Does anyone have any ideas on how to answer this question?
 A: It's a true statement.
Your statement is a universal statement, so to verify it, you can think of it as 'lining up' all objects to which the statement pertains, and checking whether they have the indicated property. So, for example, if your domain consists of two objects $a$ and $b$, then verifying the claim that all objects have property $P$ amounts to verifying that both $a$ and $b$ have property $P$. 
Ok, but what if your domain is empty (as in your case)? Well, then the verification is trivial: Yes, 'all' (zero!) objects have the property $P$. 
Here is another way to think about it: what would make a universal claim false? Well, the claim is false when you find some object that does not have property $P$ ... but if you find no such counterexamples, then the claim is true. Having no objects in your domain, however, means that there can be no counterexamples, so no matter what $P$ is, even if (as in your case) it expresses something impossible.
Consider this somewhat more concrete example:
"Every time I played the lottery, I won the jackpot!"
Huh! Am I a very lucky person? Well, consider the fact that actually I have never played the lottery ...  does that make the statement true? Yes! All zero times that I played the lottery, I won the jackpot! Again, think of it as 'lining up' all points in time when I played the lottery and verifying that at all those times I won the jackpot. Well, given that there are no such points in time where I played the lottery, this verification is trivial: yes, it checks out!
Or, in terms of counterexamples: there is no point in time where i played the lottery and did not win the jackpot. So, the claim checks out and is therefore true.
Of course, it is also true that all zero times I played the lottery, I did not win the jackpot. So, all zero times that I played the lottery, I did win the jackpot as well as did not win the jackpot ... which is just like your statement.
A: The formula is equivalent to :

$\forall x \ (x \in \emptyset \to (P(x) \land \lnot P(x)))$.

Thus, it is TRUE,because $x \in \emptyset$ is always FALSE and the truth table for the conditional has that :

FALSE $\to$ FALSE is TRUE.

See also Vacuous truth :

a statement that asserts that all members of the empty set have a certain property.

A: $$\forall x \in \emptyset. (P(x) \land \lnot P(x))$$
is true as you mentioned. 
$$(P(x) \land \lnot P(x))$$ is also true for all $x$ in the empty set.
Thus there is no conflict. 
A: Make use of the fact that for any propositions $A$ and $B$, we have $A \implies [\neg A \implies B]$ regardless of whether $B$ is true or false. This tautology is the so-called principle of explosion.
Definition: $\forall a: \neg [a\in \emptyset]$
Suppose $x\in \emptyset$ (the initial premise). 
From the definition of $\emptyset$, we obtain $\neg [x\in \emptyset]$.
Applying the principle of explosion to the initial premise, we have $\neg [x\in \emptyset] \implies P(x) \land \neg P(x)$.
Applying the rule of detachment, we have $P(x) \land \neg P(x)$. 
Discharging the initial premise and generalizing on $x$, we obtain, as required:
$\forall a:[a\in \emptyset \implies P(a)\land \neg P(a)] $
