Propositional Logic: How to reason about the "empty" list? Suppose $n=0$, how do this definition make sense in that case ? In this case the book states that this list of formulas $\Phi_i$ evalute to $\top$ for all valuations. But what reasoning can I use to prove this property of the empty list using Propositional Logic ? 
I understand, I can't use quantifiers in Propositional Logic.

 A: We say that :

$\Gamma$ tautologically implies (or logically implies or semantically entails) $\varphi$ (written $\Gamma \vDash \varphi$) iff every truth assignment for the sentence symbols in $\Gamma$ and $\varphi$ that satisfies every member of $\Gamma$ also satisfies $\varphi$.

We say that :

$\varphi$ is a tautology (written $\vDash \varphi$), when $\emptyset \vDash \varphi$, i.e. every truth assignment (for the sentence symbols in $\varphi$) satisfies $\varphi$.


In order to appreciate the first definition, we can transform it into a sort of "procedure" : 
(i) consider a valuation $v$; 
(ii) if $v$ does not satisfy some formula in $\Gamma$, throw it away;
(iii) if $v$ does, then check if it satisfy also $\varphi$.
Now it must be more evident the role of $\Gamma$ : it acts as a "filter", selecting from the set of all valuations a subset to be used for checking the satisfiability of $\varphi$.
What happens when $\Gamma = \emptyset$ ? Being empty, the emptyset cannot be used to "filter" any valuations, and thus the above "procedure" boils down to :
(i) consider a valuation $v$; 
(ii) check if it satisfy $\varphi$,
and this is exactly what the second definition amounts to.
A: I think you're asking, "When do we have '$\emptyset\models\psi$'?"
If this is the case, the answer is: only when $\psi$ is a tautology. By definition, $\emptyset\models\psi$ iff every valuation making every formula in $\emptyset$ true, makes $\psi$ true. However, every valuation at all makes every formula in $\emptyset$ true, since there aren't any. So $\emptyset\models\psi$ if and only if $\psi$ is true under every valuation.

You mention that propositional logic doesn't use quantifiers; I think this is an instance of confusing the theory and the metatheory. When we reason about propositional logic, we do so in a much more complicated logical system, where - among other things - we are allowed to use quantifiers. Definitions like that of "$\models$" take place in this metatheory.
