First order logic - is this a valid interpretation of a sentence? In the book I'm reading, the following argument is presented
There are liars. Therefore, there are thieves.
The premise is symbolised as ∃x(Lx) while the conclusion is symbolised as ∃x(Tx). The following interpretation is given to show that the argument is invalid:
Let the domain be {1,3} and the predicates L(x) and T(x) be interpreted as :
L(x) ↔ x = 1
T(x) ↔ x = 2
So that the premise ∃x(Lx) is true when x = 1 whereas the conclusion ∃x(Tx) is false with respect to the domain.
My confusion comes from '2' being in the predicate T(x). 2 isn't in the domain and I'd like to know if interpretations such as this is really allowed in first order logic. Sure the interpretation 'x=2' has the same number of free variables as T(x), but since 2 isn't in the domain, shouldn't 'x=2' technically have no meaning in this domain and therefore no truth value?
 A: You are right. Normally, in a structure with domain $\mathcal{A}$ and $\mathcal{I}$, for an $n$-ary predicate $P$, $\mathcal{I}(P^n) \subseteq \mathcal{A}^n$, i.e. interpretation of an 1-ary predicate is a subset of the domain, the interprtation of an $2$-ary predicate is a subset of $\mathcal{A} \times \mathcal{A}$, ...,
Your book might presuppose some weird definition of models in which non-logical symbols can be interpreted as anything, but normally you'd want models to be closed systems, in the sense that interpretating some predicate shouldn't shoot you out of the domain of the model.  
In order to prove that $\exists x L(x) \not \vDash \exists x T(x)$, they should rather have presented a model in which there are liears but no thieves at all, e.g.
$$\text{domain} = \{1\};\ \mathcal{I}(L) = \{1\};\ \mathcal{I}(T) = \emptyset$$
Or, with the $\leftrightarrow$ notation,
$$L(x) \leftrightarrow x = 1;\ T(x) \leftrightarrow \bot$$
- but this notation presupposes that 1 (and 2 etc.) are constants, which, as Mauro Allegranza points out in their comment, need to be assigned an interpretation as well in order to be any meaningful.
May I ask which book you are using, and how they define models? If they nowhere give a precise definition of what an interpretation of a predicate is, then it's probably not a good book anyway.  
A: Your counter-example with domain $D = \{ 1,3 \}$ can be misleading, due to the fact that the number two is not in $D$...
It works when we interpret $L(x)$ as the property "x is equal to one" and $T(x)$ as the property "x is equal to two".
We have to carefully avoid the conflation of "objects" and "names": according to the semantics for first-order logic we can have objects (i.e. elements of the domain) without name (i.e. without individual constants referring to them) but we cannot have constants without reference.
Another counter-example can be based on the following interpretation for $L(x)$ and $T(x)$ respectively :

"$x \text { is Odd}$" and "$x \text { is Even}$".

In this case, the argument will become:

There are Odd numbers. Therefore, there are Even numbers.

which is clearly falsified in $\{ 1,3 \}$.
But, IMO, the examples above are not based on what the author call: arithmetical interpretation, i.e. an interpretation in the domain of positive integers (see Patrick Suppes, Introduction to Logic, page 64).
An arithmetical counter-example, based on domain $\mathbb N = \{ 1, 2, \ldots \}$, will be the following:

let $L(x)$ interpreted as $(x \ge 1)$ and let $T(x)$ interpreted as $(x < 1)$.


Having found suitable interpretations showing that: $ ∃xL(x) \nvDash  ∃xT(x)$, we have showed that the argument:

There are liars. Therefore, there are thieves.

is not valid. 
A: When it says:

Let the domain be {1,3} and the predicates L(x) and T(x) be interpreted as :
L(x) ↔ x = 1
T(x) ↔ x = 2

it is not considering $x=2$ to be a logic formula that is yet to be interpreted, but rather it indicates for which objects of the domain the predicate holds true (so yes, that's very confusing; they mix up logic notation with mathematical expressions about the interpretation)
So, it says that the predicate $T$ holds for object $2$ ... which is not part of the domain .. and so in effect there are no thieves at all.  Which is what you want, since object $1$ is a liar, and therefore the premise is true, and the conclusion is false, and thus we have a counterexample, as desired.
What I don;t understand, though, is that they could simply have picked the domain as $\{ 1 \}$ ... that would have worked just as well.
