Tautological and logical consequence In Herbert Enderton's book A Mathematical Introduction to Logic, it is mentioned [see page 115] that $Pc$ is not a tautological consequence of $\forall xPx$ (when both are taken as sentence variables for propositional calculus) but $Pc$ is a logical consequence of $\forall xPx$.
Suppose we let variables $u$ and $v$ (belonging to $\{ 0,1 \}$) represent the truth values of $\forall xPx$ and $Pc$ respectively.
Now "$Pc$ is not a tautological consequence of $\forall xPx$ (and vice versa)" would be equivalent to saying that "variables $u$ and $v$ are independent of each other".
On the other hand, "$Pc$ is a logical consequence of $\forall xPx$" is equivalent to saying "variables $u$ and $v$ are not independent of each other" [in fact, it is saying ,"$u \le v$"]. 
Since both "$Pc$ is not a tautological consequence of $\forall xPx$" and 
"$Pc$ is a logical consequence of $\forall xPx$" are true, where is the flaw in the above reasoning?
 A: To say that $B$ is a tautological consequence of $A$ is in effect to say that $B$ is a logical consequence of $A$ in virtue of the distribution of truth-functional connectives in $A$ and $B$ (as we might put it, $A$ entails $B$ by the truth-table test). That's why $Pc$ is not a tautological consequence of $\forall xPx$ -- certainly $\forall xPx$ logically entails $Pc$, but plainly not in virtue of the distribution of truth-functional connectives in the premiss and the conclusion.
The OP writes

"$Pc$ is not a tautological consequence of $\forall xPx$  ( and vice versa )"would be equivalent to saying that "[the truth values of $Pc$ and $\forall xPx$] are independent of each other". 

Not so. To say "$Pc$ is not a tautological consequence of $\forall xPx$" only implies that that the logical relation between the propositions, whatever it is, is not settled by the distribution of truth-functional connectives in them.
A: The basic difference between tautological consequence and logical consequence is related to the different "expressive capabilities" of sentential logic language compared to first-order one.
Consider a tautology like : $A \lor \lnot A$. 
Every instance of it is true; thus, if we consider the instance obtained replacing the sentential letter $A$ with a f-o logic formula, like : $\forall xP(x)$, we still have a tautology :

$\forall xP(x) \lor \lnot \forall xP(x)$.

As you can see in Enderton's book [see page 115 and Exercises 3, page 129] :

If $\Gamma$ tautologically implies $\varphi$, then it follows that $\Gamma$ also logically implies $\varphi$. But the converse fails.

Consequently, every tautology is valid, but it is not true that every valid formula is a tautology.
Enderton's example [see page 115] discussed above amounts to show this fact.
For simplicity, I'll rewrite it using "tautology" and "valid", exploiting the fact that [see Enderton, page 99] : 

$\alpha \vDash \varphi$ iff  $\vDash \alpha \rightarrow \varphi$.

Consider now the f-o formula : $Pc \rightarrow \forall xP(x)$.
If we "read it" as a formula of sentential logic [see discussion at page 114], it is composed of two distinct sub-formulae; thus we have to translate it using two different sentential letters.
What we get is : $A \rightarrow B$, which is not a tautology.
As you said, we can choose a truth assignment $v$ such that $v(A)=1$ and $v(B)=0$ and the result will be : $v(A \rightarrow B)=0$, exactly because the two sentential letters are independent.
Recall my first statement about the different "expressive capability" of the two languages : sentential logic is not able "to see" the interdependency between $Pc$ and $\forall xP(x)$.
Of course, the same argument applies to : $\forall xP(x) \rightarrow Pc$.


Consider now the f-o case; we need to extend the rules for "calculating" truth assignments.
We still have :


*

*$v(\lnot \varphi) = 1 - v(\varphi)$


and :


*

*$v(\varphi \rightarrow \psi) = max(1-v(\varphi), v(\psi))$


but we need also a rule for the universal quantifier ($\forall$) :


*

*$v(\forall x P(x))=inf [v(P(t/x))]$, where inf is "calculetad" taking into account all $t$.


Again, we say that $\alpha$ is valid iff $v(\alpha)=1$, for all $v$.

a) : $Pc \rightarrow \forall xP(x)$
Consider an interpretation of the above formula in a domain where $P$ holds of $c$, but assume also that there are some elements in the domain such that $P$ does not hold of them.
[We can think at the set $\mathbb N$ of natural numbers as domain of the interpretation and interpret the predicate letter $P$ with the property $Odd(x)$, and assume that the constant $c$ denote the number $1$. Clearly $Odd(1)$ is true, but $Odd(2)$ is not.]
Using this interpretation to define a truth assignment $v$, we have that :
$v(Pc)=1$, while $v(P(t/x))=0$, for some $t$; 
thus : 
$v(\forall xP(x))=inf [v(P(t/x))]=0$.
In conclusion : 

$v(Pc \rightarrow \forall xP(x))=max[1-v(Pc), v(\forall xP(x))]=max(0,0)=0$

and we conclude that the formula is not valid, and consequently :


$$Pc \nvDash \forall xP(x)$$



b) : $\forall xP(x) \rightarrow Pc$
Now we have two cases to consider :
$b_1$) : $P$ holds of every element in the domain of the interpretation. Thus, it holds also of $c$.
[Reusing the set $\mathbb N$, now we can interpret the predicate letter $P$ with the property $x \ge 0$, and again assume that the constant $c$ denote the number $1$. Thus : $\forall x(x \ge 0)$ and $1 \ge 0$ are both true.]
This implies : $v(P(t/x))=1$, for all $t$, and so also : $v(Pc)=1$. 
Thus, $v(\forall xP(x))=inf [v(P(t/x))]=1$, and we have :

$v(\forall xP(x) \rightarrow Pc) = max[1-v(\forall xP(x)), v(Pc)]=max(0,1)=1$.

Note : as you ca see, it is in this case that the truth values assigned by $v$ to $Pc$ and $\forall xP(x)$ are not independent of each other.

$b_2$) : it is not true that $P$ holds of every element in the domain of the interpretation. 
[Consider now $\mathbb N$ with the property $x > 0$; $\forall x(x > 0)$ is not true, because $0 \le 0$.]
This implies : $v(P(t/x))=0$, for some $t$, and thus :
$v(\forall xP(x))=inf [v(P(t/x))]=0$.
Now, it does not matter whether $P$ holds of $c$ or not; we can calculate :

$v(\forall xP(x) \rightarrow Pc) = max[1-0, v(Pc)]=max(1,v(Pc))=1$.

In both cases $v(\forall xP(x) \rightarrow Pc) = 1$ and we conclude that the formula is  valid, and consequently :


$$\forall xP(x) \vDash Pc$$


