Help with truth statements "For all $x \in \varnothing, p(x)$" and "$\exists x \in \varnothing, p(x)$ " Why is the statement "For all $x \in \varnothing, p(x)$" always true, and why is the statement "$\exists x \in \varnothing, p(x)$ always false for a statement $p(x)$ that depends on $x$?
 A: $\exists x\in\emptyset~.p(x)$ must be false, because there is no element in the emptyset that could satisfy predicate $p$.
$\exists x\in\emptyset~.\lnot p(x)$ must be false, because there is no element in the emptyset that could not satisfy predicate $p$.
So $\forall x\in\emptyset~.p(x)$ must be true, because there is no element in the emptyset that could not satisfy predicate $p$.
A: The proposition $\forall x \in \emptyset, p\left(x\right)$, in its formal form is:
$$\forall x, x \in \emptyset \rightarrow p\left(x\right).$$
As $x \in \emptyset$ is always false, the statement $x \in \emptyset \rightarrow p\left(x\right)$ is always true. In fact, You can think of $p \rightarrow q$ as equivalent to $\left(\lnot p\right) \vee q$.
On the other hand, the proposition $\exists x \in \emptyset, p\left(x\right)$ is formally written as
$$\exists x, x \in \emptyset \wedge p\left(x\right).$$
As $x \in \emptyset$ is always false, so is $x \in \emptyset \wedge p\left(x\right)$. As a result, the proposition $\exists x, x \in \emptyset \wedge p\left(x\right)$ can never be accepted.
Strictly speaking, what I have explained is not formal proof. It is based on intuitive semantics instead of rigorous syntax. The proof for these two propositions requires using deduction rules, which is another story in first order logic.
A: I think this case is easiest to understand if we consider the quantifier $\exists$ first, and then define $\forall$ in terms of $\exists$.
Let $S$ be a non-empty set.
We can make the following claim and justify it intuitively.
$$ \exists x : S \mathop. \varphi(x) \;\;\text{if and only if}\;\; \{ x :x\in S \land \varphi(x) \} \ne \emptyset $$
In prose, (there exists an $x$ in S such that $\varphi(x)$) if and only if (the set of all $x$s such that $x$ is in $S$ and $\varphi(x)$ is nonempty).
Let's extend this formula to the case where $S$ is empty.
$$ \{ x : x \in \emptyset \land \varphi(x) \} \ne \emptyset $$
However, $x \in \emptyset$ is never true, therefore this expression simplifies to
$$ \emptyset \ne \emptyset $$
which is false. Thus
$$ \exists x : \emptyset \mathop. \varphi(x) \;\;\text{is never true}$$
Next, we think about what forall means, then we can define it in terms of existence.
$$ \forall x : S \mathop. \varphi(x)  \;\;\text{if and only if}\;\;  \lnot \exists x : S \mathop. \lnot \varphi(x)$$
In prose, $\varphi$ is true for all $x$ in $S$ if and only if there does not exist an $x$ in $S$ such that $\varphi$.
If we apply this definition, then we get that forall, when quantified over the empty set, is true.
$$ \forall x : S \mathop. \varphi(x) $$
$$ \lnot \exists x : S \mathop. \lnot \varphi(x) $$
$$ \lnot \bot $$
$$ \top $$
A: Consider this statement : I am older than my father & I have a car. Is this true?
Well, you do not know whether I have a car ( in fact I have one) but you can surely know that I'm not older than my father, since it is impossible. Being given the whole " and" statement contains an impossible conjunct, this whole statement must be false.
In the same way : it is impossible that $\exists (x)\space [ x\in \emptyset]$
so, a fortiori, it is impossible that $\exists (x) \space [ x\in \emptyset \space \& \space \phi(x)] $   (whatever the property  $\phi$ may be).
And, since what is impossibly true just isn't true : it is false that $\exists (x) \space [ x\in \emptyset \space \&\space  \phi(x)\space] $


*

*The statement "   $\forall (x)_{\in \emptyset}, \phi(x)$ " is a conditional, namely, the conditional : " $\forall(x) [ x\in\emptyset \rightarrow\phi(x) ] $".


*Let's drop the quantifier and only consider the " if... then" statement inside, that is :

$x\in\emptyset \rightarrow\phi(x)$

and let's try to find an object $x$ that would make the statement false . So consider whatever object you want: number 5, The Moon, The Eiffel Tower, whatever, and let this object play the role of $x$ in the conditional statement.

*

*What do we notice? That in each case, the antecedent of the " if ... then " is false ( for, of course, it is always false to say that an object belongs to the empty set, since it is impossible ).


*Now, look at the truth table of the " if ... then " operator.


*

*You can see that, whatever object you choose to play the role of " x" , you are on line 3 or on line 4 of the truth table. And on these lines, the " if ... then " is always true .


*Note that here the truth value of the consequent  "$\phi(x)$ has no impact on the truth value of the whole conditional.


*Conclusion : since what we discovered holds for any $x$, it also holds for all $x$, hence the statement

$\forall(x) [ x\in\emptyset \rightarrow\phi(x) ] $

