Are truth tables valid for universal statements? Why or why not? I'm reading the book on Discrete Mathematics by Kevin Ferland. In Section 1.5, he says truth tables are not an option for statements involving universal quantifiers.
It seems to me that a statement such as "$\forall\ x \in \mathcal{U}, p(x)$" is either true or false always. I can think of examples: "Any rational number is real." This is a universal statement, and it is true. 
Is it possible to choose a universal set $\mathcal{U}$ and condition $p$ so that "$\forall\ x \in \mathcal{U}, p(x)$" does not evaluate to a logical statement (something that is either true or false, but not both)? If so, can you please give a concrete example? If not, why then can I not create a truth table, and going further, why can't I validate argument forms using the truth table? 
As an example, let's try to validate the following argument form using a truth table. 
\begin{align*}
\forall\ x \in \mathcal{U}, p(x)\vee q(x)\\
\forall\ x \in \mathcal{U}, \neg p(x)\\
\therefore, \forall\ x \in \mathcal{U}, q(x)
\end{align*}
So, we need to show that whenever the premises are both true, the conclusion is also true. Since it's hard to write $\forall$ in the table, assume each of the columns is preceded by $\forall\ x \in \mathcal{U}$. In other words, each statement is universal.
p(x)  q(x)  p(x) v q(x)  ~p(x)  q(x)
----+-----+------------+------+-----
  T |  T  |    T       |   F  |  T
  T |  F  |    T       |   F  |  F
  F |  T  |    T       |   T  |  T
  F |  F  |    F       |   T  |  F

Now, whenever the premises are both true, the conclusion is true. So one would think that the argument form is valid. And indeed, it is valid. You can certainly use universal instantiation and proofs of basic argument forms (which CAN be validated with truth tables) to show that the argument form is valid. This is the way that makes sense to me. 
What confuses me is why is the above method flawed? (In general).
Any explicit examples you can give me would greatly help! Thanks in advance.
 A: Suppose $\mathcal{U}$ was some arbitrary set of infinite cardinality. 
I think the issue that the book is touching on is that for some arbitrary statement containing a universal quantifier (like $\forall x \in \mathcal{U} \ p(x)$), although it does have a truth value, you cannot use a truth table to find that value directly by testing all values of $x$. (since you would have to have an infinitely large table to get all of the cases)
What made your example doable with a truth table was that in that particular case, you did not have to consider every possible $x$ (which could be infinite), you only needed to care about $p(x), q(x), p(x) \vee q(x)$, which can only take on finitely many values.
A: The only problem I see with this is that it seems you might think it's okay for $\mathcal{U}$
to be the universe.
Are you in fact saying that?
What do you mean by a "universal set?"
If instead $\mathcal{U}$
is some other set guaranteed by the axioms of ZFC,
then the axiom of specification guarantees that for any formula $p$
there exists a set $S=\{x\in\mathcal{U}\mid p(x)\}$.
It is then either true or false that $S=\mathcal{U}$,
and the trueness or falsehood if this equation is equivalent to that of your statement $\forall x\in\mathcal{U}:p(x)$.
Upon making the desired truth table then, we are merely considering the truth values that result from varying $p$ and $q$, not from varying $x$.
This puts us in exactly the same situation as we would be for an ordinary truth table, thus I don't see any problem. You just have to write it correctly.
A: Remember that FOL is stronger than propositional logic.  Ie. there are statements in FOL that can not be expressed in propositional logic.  Your statement is not one of these.
Consider
\begin{align*}
\forall\ x \in \mathcal{U}, p(x)\vee q(x)\\
\neg p(a)\vee\forall\ x\in\mathcal{U}, q(x)\\
\therefore q(a)
\end{align*}
What does your thruth table look like?
