Is this a tautology: $\forall xP(x) \implies Q(x)$ if there's no $x$ such that $P(x)$? Is this a tautology: $\forall xP(x) \implies Q(x)$ if there's no $x$ such that $P(x)$?
I know that if there was an exists there instead of a for all, the antecedent would be false and thus the implication true. But this one confuses me. 
Could someone clarify this form me? 
E: This question comes from an exercise I have which says: Let $\Gamma$ be an unsatisfiable set of formulas. Let $\alpha$ be a formula. Does $\alpha\in C(\Gamma) $, where $C(X)$ denotes the consequences of $X$?
The definition of consequence I have says $\alpha\in C(X)$ if $\forall v$ valuation$:v(\Gamma)=\{1\} \implies v(\alpha)=1$. As my $\Gamma$ has no valuations that satisfy it, this should be true, right?
Thanks!
 A: The formula is a predicate (or first-order) one: in this context, we have two uses of the term tautology :
(i) a f-o formula is called a tautology when it is an "instance" of a propositional tautology. The formula $∀xP(x) \to Q(x)$ is an instance of the propositional formula $p \to q$, which is not a tautology.
(ii) a f-o formula is (sometimes) called a tautology when it is valid (i.e. true in every interpretation). The formula $∀xP(x) \to Q(x)$ is - as you say - true in every interpretation where there are no $P$'s, but of course is not valid.
To verify this, we can interpret the formula in $\mathbb N$ and interpret $Q$ with $x < 0$ and $P$ with $x \ge 0$ : clearly $\forall x (x \ge 0) \to (x < 0)$ is false in $\mathbb N$.

Thus, in both cases, it is not correct to say that the formula is a tautology.
In all mathematical logic textbooks (see e.g. Enderton, page 23) a formula $\varphi$ of propositional logic is defiend as a tautology (written $\vDash \varphi$) when:

every truth assignment (for the sentence symbols in $\varphi$) satisfies it.

If we "extrapolate" this definition to the first-order language, we get the definition of valid formula:

we say that $\varphi$ is valid when every interpretation satisfy it,

and this is not the case of the formula above.
What we have is : $¬∃P(x) \vDash ∀xP(x) \to Q(x)$, that reads : 

$¬∃P(x)$ logically (and not : tautologically) implies $∀xP(x) \to Q(x)$.


Regarding the added part, the answer is : YES.
If no truth assignment satisfies every member of $\Gamma$, then for any $\varphi$, it is vacuously true that $\Gamma \vDash \varphi$.
A: 
Is this a tautology: $\forall x[P(x) \implies Q(x)]$ if there's no $x$ such that $P(x)$?  (Inserted presumed brackets.)

Yes. 
Disclaimer: This proof probably won't be using exactly the axioms you are allowed, but it should be possible to translate it into your own system.


*

*We have $\forall x: \neg P(x)$

*Suppose $P(y)$.

*From 1, we have $\neg P(y)$.

*From 3, we have $\neg\neg P(y)\implies Q(y)$ since anything follows
from a falsehood.

*From 4, $P(y)\implies Q(y)$, removing the $\neg\neg$. 

*From 2 and 5, $Q(y)$ by detachment  

*From 2 and 6, $P(y)\implies Q(y)$, the conclusion. 

*Generalizing, $\forall x :[P(x)\implies Q(x)] $
