Interpreting Quantified Predicate Formulas Does $\forall x (P(x) \rightarrow Q(x))$ mean that $(P(x) \rightarrow Q(x))$ is true for all truth assignments to $x$ or does it mean that  $P(x)$ and $Q(x)$ can be true or false depending on the truth assignment to $x$?
 A: I think it should come with a domain for $x$.
With a domain for $x$, if this statement is true, it means "For every $x$ in the domain, if $P(x)$ is true, then $Q(x)$ is true; if $P(x)$ is false, then $Q(x)$ is false".
The statement is false $iff$ there exists $x$ such that $P(x)$ is true but $Q(x)$ is false.
A: Formulas of first-order (or predicate) logic, like :

$∀x(P(x)→Q(x))$

are interpeted not with truth assignments, but through an interpretation $\mathcal M$, made of :


*

*a (not empty) set of objects, called the domain of the interpretation; call it $M$


*a suitable reference for every constant of the language (if any)


*an assignement to each predicate letter of the language of a relation on $M$ of the suitable arity; i.e. if the predicate letter $P$ (as in our example) is monadic (i.e.has only one argument: arity = $1$) then we assign to it a subset $P^M$ of M (it will be the set of objects of the domain for which $P$ holds); if $R$ is a binary relation (a predicate with arity = $2$) then we assign to it a sucset $R^M$ of $M \times M$; and so on.

Consider now the language of arithmetic, with an individual constant : $\overline 0$, a function symbol $S$ for the successor function and a binary relation : $<$ for less than.
We interpret it in the domain $\mathbb N$ of natural numbers, with the obvious interpretation for the above symbols.
Consider now the formula :

$\forall x(x = \overline 0)$.

Intuitively, we expect that it will be true in $\mathbb N$ if, substituting to the individual variable the name for an object in the domain $\mathbb N$, we obtain a true formula, and this will happen with all numbers, i.e. for every object in the domain.
Consider a formula $A(x)$ with only $x$ free; we call an instance of it a formula $A[t/x]$ obtained from $A(x)$ putting the closed term $t$ (i.e. a "name" of an object) in place of $x$.
Assume that we have a name for every number (for every object in $\mathbb N$); we use $\overline k$ as name for the number $k$.
Now it's done !


The formula $\forall x (x = \overline 0)$ is true in $\mathbb N$ iff every instance : $(x = \overline 0)[\overline k/x]$ is true in $\mathbb N$.


With $0$ as $k$, we have that $(x = \overline 0)[\overline 0/x]$ is simply : $\overline 0 = \overline 0$ wich is true.
But with $1$ as $k$, we have that $(x = \overline 0)[\overline 1/x]$ is : $\overline 1 = \overline 0$ wich is obviously false.
Thus, having found an instance of $x = \overline 0$ which is false, we may conclude that (as expected) :


$\forall x (x = \overline 0)$ is false in $\mathbb N$.



Now, what about

$∀x(P(x)→Q(x))$ ?

First of all, we have to choose a domain $M$ for our interpretation $\mathcal M$.
Second, we have to assign to the predicate letters $P$ and $Q$ two subset of the domain $M$, $P^M$ and $Q^M$ respectively.
Having done this, we have that the above formula will be true in $\mathcal M$ iff for every $a \in M$, if $\overline a$ is the name of $a$, the formula $(P(x)→Q(x))[\overline a/x]$ is true in $\mathcal M$.
In other terms,


$∀x(P(x)→Q(x))$ is true in the interpretation $\mathcal M$ iff for every $a \in M$, $(P(\overline a)→Q(\overline a))$ is true in it.


