Domain of discourse and quantifying in predicate logic I am struggling with an idea about how quantifiers relate to domains of discourse.
Given a statement "$x$ is divisible by $2$" represented by the predicate $D(x)$,the predicate currently has no truth value since it contains only a free variable. But if I define the domain to be all even numbers I have essentially given $D(x)$ a truth value without binding the variable with a quantifier. Why is is this so? I'm assuming that by defining my domain I somehow introduced some kind of binding quantification on the statement. Since $x$ is still a free variable how can the statement have a truth value?
 A: It's not uncommon for logics to assume universal quantification on all unbound variables.  ACL2 does this, for example.  
However, most logics also assume some predefined collection of constants.  In some computer logics that set is something similar to all real numbers, integers, strings, all lists of the preceding.
The Universe of a logic isn't something that is defined while using the logic, it is something that is defined while creating the logic.  If you define your logical constants to be $0, 2, 4, 6, 8....$ then calling the constants "even numbers" is a different meaning than what calling $D(x)$ even numbers is.  The first is a statement about how the logic is created in terms of some more universal idea, the second is a statement defined using the logic itself.
It is actually quite nonsensical to "change your domain of discourse", because in doing so you have defined a new logic, and then you have to redefine $D(x)$ in the new logic, using the new set of constants available to you.
I'm not aware of any logic that admits removals of objects from the Universe as a method of using the logic.  However, some logics do allow for the introduction of new objects with varying degrees of usefulness.
A: "But if I define the domain to be all even numbers I have essentially given D(x) a truth value without binding the variable with a quantifier."
This isn't quite right. What you've really done is find a model that makes the sentence $(\forall x)(Dx)$ true. If you add a single odd integer to your domain of discourse, the sentence $(\forall x)(Dx)$ becomes false. Truth values are relative to models.
In any case, $Dx$, an open formula, does not have a truth value, no matter what model you work with. Only sentences, aka closed formulas (formulas in which all the variables are bound) have truth values in a model. Rather, in a model, $Dx$ stands for a truth function: its truth value is a function of the variable $x$; it acquires a specific truth value only by substituting a value for the variable $x$.
A: 
I am struggling with an idea about how quantifiers relate to domains of discourse. Given a statement "x is divisible by 2" represented by the predicate D(x),the predicate currently has no truth value since it contains only a free variable.

Some will disagree, but I'm not sure that the notions of domain of discourse or truth-value are all that useful in this context. You can start a proof with an initial premise $D(x)$ whether you have defined $D$ or not, and draw simple inferences from it. 
Example:


*

*$D(x)$  (Assumption)

*$D(x) \land D(x)$ (Intro $\land,$ 1, 1)

*$D(x)\implies D(x) \land D(x) $ (Intro $\implies$, 1, 2)

*$\forall a:[D(a)\implies D(a)\land D(a)]$ (Intro $\forall$, 3)
In this statement, we are quantifying over all objects, real or imagined. There is no domain of discourse here as such.
Let's suppose that you had previously defined an "is divisible by 2" predicate $D$ as follows:
$\forall a\in N: [D(a)\iff \exists b\in N: a=2b]$
To apply this definition for any object $x$, you would previously have had to assume (or prove) that $x\in N.$ You would not be able to apply this definition for $x$ if you assumed only that $D(x)$ is true. 
